Would adding water decrease the forward and reverse reaction rates of a system in equilibrium?
Q. It would be diluting the system. Thus, less collisions between reactant/product particles occur per unit of time, thus decreasing forward and reverse reaction rates. Is this correct? By the way, the water is in a liquid state. Thanks in advance. 10 points for best answer. It keeps posting twice for some reason.
Asked by AFL Fanatic - Sun Oct 4 02:44:08 2009 - - 2 Answers - 0 Comments
A. Generally, adding water should decrease the reaction rate. In some cases, as in overly concentrated systems, it seems that adding water actually can increase the reaction rate - likely with certain biochemical situations. This may have something to do with decreasing chemical interferences. Absent complex biochemical matrixes, adding water to a simple chemical system should decrease the rate of reaction, fo rhe reasons you set forth.
Answered by jethro - Mon Oct 5 06:55:24 2009
Q. It would be diluting the system. Thus, less collisions between reactant/product particles occur per unit of time, thus decreasing forward and reverse reaction rates. Is this correct? By the way, the water is in a liquid state. Thanks in advance. 10 points for best answer. It keeps posting twice for some reason.
Asked by AFL Fanatic - Sun Oct 4 02:44:08 2009 - - 2 Answers - 0 Comments
A. Generally, adding water should decrease the reaction rate. In some cases, as in overly concentrated systems, it seems that adding water actually can increase the reaction rate - likely with certain biochemical situations. This may have something to do with decreasing chemical interferences. Absent complex biochemical matrixes, adding water to a simple chemical system should decrease the rate of reaction, fo rhe reasons you set forth.
Answered by jethro - Mon Oct 5 06:55:24 2009
Would u expect the rate constant for reverse reaction 2b smaller or larger than that for the forward reaction?
Q. GIVEN: The forward reaction HBr(g) + Br(g) H(g) + Br2S(g) is endothermic... can you explain briefly.? thanks alot
Asked by ryan - Thu Jun 26 18:14:32 2008 - - 1 Answers - 0 Comments
A. Where the element S is from in "Br2S"?
Answered by Hahaha - Thu Jun 26 22:06:10 2008
Q. GIVEN: The forward reaction HBr(g) + Br(g) H(g) + Br2S(g) is endothermic... can you explain briefly.? thanks alot
Asked by ryan - Thu Jun 26 18:14:32 2008 - - 1 Answers - 0 Comments
A. Where the element S is from in "Br2S"?
Answered by Hahaha - Thu Jun 26 22:06:10 2008
when a reaction is at equilibrium and more reactant is added which of the following changes is the?
Q. immediare result? A. the reverse reaction rate remains the same B. the forward reaction rate increases C. the reverse reaction rate decreases D. the forward reaction rate remains the same
Asked by Ariel - Sun Apr 26 00:19:32 2009 - - 2 Answers - 0 Comments
Q. immediare result? A. the reverse reaction rate remains the same B. the forward reaction rate increases C. the reverse reaction rate decreases D. the forward reaction rate remains the same
Asked by Ariel - Sun Apr 26 00:19:32 2009 - - 2 Answers - 0 Comments
Which is a property of a reaction that has reached equilibrium?
Q. The rate of the forward reaction is greater than the rate of the reverse reaction. The rate of the forward reaction is equal to than the rate of the reverse reaction. The amount of products is equal to the amount of reactants. The amount of products is greater than the amount of reactants.
Asked by Marley J. - Tue Mar 25 13:41:17 2008 - - 1 Answers - 0 Comments
A. B
Answered by rockyfella25 - Tue Mar 25 13:45:45 2008
Q. The rate of the forward reaction is greater than the rate of the reverse reaction. The rate of the forward reaction is equal to than the rate of the reverse reaction. The amount of products is equal to the amount of reactants. The amount of products is greater than the amount of reactants.
Asked by Marley J. - Tue Mar 25 13:41:17 2008 - - 1 Answers - 0 Comments
A. B
Answered by rockyfella25 - Tue Mar 25 13:45:45 2008
What is the new value of the reverse reaction constant, k(r), after adding catalyst? (see details)?
Q. Subscripts in parenthesis ( ) For a different reaction, K(c) = 4.46, K(f) = 111 s^-1, and k(r) = 24.9s^-1. Adding a catalyst increases the forward rate constant to 7.66 10^3 s^-1. What is the new value of the reverse reaction constant,K(r) , after adding catalyst?
Asked by pikachuflatulates - Sun Mar 8 23:12:22 2009 - - 1 Answers - 0 Comments
A. Catalyst does not change the value of the equilibrium constant K(c) So the relation K(c) = k(f)/k(r) with K(c) = 4.46 holds for rate constants of catalyzed reaction Hence: k(r) = k(f) / K(c) = 7.66 10 s / 4.46 = 1.74 10 s
Answered by schmiso - Thu Mar 12 17:06:22 2009
Q. Subscripts in parenthesis ( ) For a different reaction, K(c) = 4.46, K(f) = 111 s^-1, and k(r) = 24.9s^-1. Adding a catalyst increases the forward rate constant to 7.66 10^3 s^-1. What is the new value of the reverse reaction constant,K(r) , after adding catalyst?
Asked by pikachuflatulates - Sun Mar 8 23:12:22 2009 - - 1 Answers - 0 Comments
A. Catalyst does not change the value of the equilibrium constant K(c) So the relation K(c) = k(f)/k(r) with K(c) = 4.46 holds for rate constants of catalyzed reaction Hence: k(r) = k(f) / K(c) = 7.66 10 s / 4.46 = 1.74 10 s
Answered by schmiso - Thu Mar 12 17:06:22 2009
Which of the following describes when any reaction that has reached chemical equilibrium?
Q. A. All of the products have been converted to the reactants of the reaction. B. All of the reactants have been converted to the products of the reaction. C. It's not see :) D. Noth the forward and the reverse reactions have stopped with no net effect on the concentration of the reactants and the products. E. The rate of the forward reaction is equal to the rate of the reverse reaction. Please help. I was told to take home and correct answers I got wrong on a test and this is the only one I don't get.
Asked by itsTylerRandall - Wed Sep 16 21:05:40 2009 - - 1 Answers - 0 Comments
A. E This means on average over time, the amount of products and reactants are the same at equilibrium.
Answered by Kevin H - Wed Sep 16 21:09:33 2009
Q. A. All of the products have been converted to the reactants of the reaction. B. All of the reactants have been converted to the products of the reaction. C. It's not see :) D. Noth the forward and the reverse reactions have stopped with no net effect on the concentration of the reactants and the products. E. The rate of the forward reaction is equal to the rate of the reverse reaction. Please help. I was told to take home and correct answers I got wrong on a test and this is the only one I don't get.
Asked by itsTylerRandall - Wed Sep 16 21:05:40 2009 - - 1 Answers - 0 Comments
A. E This means on average over time, the amount of products and reactants are the same at equilibrium.
Answered by Kevin H - Wed Sep 16 21:09:33 2009
Reaction rates, easy 10 points? Yes/No question?
Q. Say the reaction was the decomposition of CaC03(s). If I added more CaCO3 solid, the forward and reverse reaction rate would not change. Correct? Thanks in advance. 10 points for best answer.
Asked by AFL Fanatic - Tue Mar 3 05:12:06 2009 - - 2 Answers - 0 Comments
A. yes
Answered by flipper3465 - Tue Mar 3 05:14:28 2009
Q. Say the reaction was the decomposition of CaC03(s). If I added more CaCO3 solid, the forward and reverse reaction rate would not change. Correct? Thanks in advance. 10 points for best answer.
Asked by AFL Fanatic - Tue Mar 3 05:12:06 2009 - - 2 Answers - 0 Comments
A. yes
Answered by flipper3465 - Tue Mar 3 05:14:28 2009
One effect of a catalyst being added to a reaction mixture is?
Q. a. to increase the equilibrium constant for the reaction b. to slow down the rate of the reverse reaction c. to raise the temperature of the mixture d. to provide a new pathway for the reaction e. none of these choices is correct
Asked by *Cookie* - Thu Apr 16 16:12:06 2009 - - 2 Answers - 0 Comments
A. The answer is D It cannot be A because catalysts do not change the equilibrium constant for a reaction. B: As far as im concerned, a catalyst may increase the speed of reaction for both reactions (direct and reversed). C: Catalysts have no effect on the enthalpy of a reaction D:Its D, as you know, catalyst speed up the speed of a specifiq reaction, they do so by reducing the by changing the Activation Energy of a specifiq reaction, they are able to do that because they change the Reaction Mecanism (or in your confusing terms, they provide a new pathway for the reaction, that is a new way of obtaining the final product) E: Its not E because D is correct
Answered by Cool-dawg - Thu Apr 16 16:21:39 2009
Q. a. to increase the equilibrium constant for the reaction b. to slow down the rate of the reverse reaction c. to raise the temperature of the mixture d. to provide a new pathway for the reaction e. none of these choices is correct
Asked by *Cookie* - Thu Apr 16 16:12:06 2009 - - 2 Answers - 0 Comments
A. The answer is D It cannot be A because catalysts do not change the equilibrium constant for a reaction. B: As far as im concerned, a catalyst may increase the speed of reaction for both reactions (direct and reversed). C: Catalysts have no effect on the enthalpy of a reaction D:Its D, as you know, catalyst speed up the speed of a specifiq reaction, they do so by reducing the by changing the Activation Energy of a specifiq reaction, they are able to do that because they change the Reaction Mecanism (or in your confusing terms, they provide a new pathway for the reaction, that is a new way of obtaining the final product) E: Its not E because D is correct
Answered by Cool-dawg - Thu Apr 16 16:21:39 2009
A chemist adds a catalyst to a reaction. What will happen to the reaction?
Q. A chemist adds a catalyst to a reaction. What will happen to the reaction? Classify the following statements as True or False. 1. The activation energy of only the forward reaction will decrease. 2. The rate of only the forward reaction will increase. 3. A stoichiometric amount of catalyst will work best. 4. The rate of the forward and reverse reactions will increase. 5. The same products will be formed as for the uncatalyzed reaction. 6. The catalyst will be regenerated at the end of the reaction.
Asked by Alex M - Fri Feb 6 08:08:51 2009 - - 4 Answers - 0 Comments
A. 1.False 2.False 3. False 4.True 5. True, although product ratio may change. Catalyst may favor some reaction more than other in parallel reactions. 6.True
Answered by See-saw Mind - Fri Feb 6 08:15:27 2009
Q. A chemist adds a catalyst to a reaction. What will happen to the reaction? Classify the following statements as True or False. 1. The activation energy of only the forward reaction will decrease. 2. The rate of only the forward reaction will increase. 3. A stoichiometric amount of catalyst will work best. 4. The rate of the forward and reverse reactions will increase. 5. The same products will be formed as for the uncatalyzed reaction. 6. The catalyst will be regenerated at the end of the reaction.
Asked by Alex M - Fri Feb 6 08:08:51 2009 - - 4 Answers - 0 Comments
A. 1.False 2.False 3. False 4.True 5. True, although product ratio may change. Catalyst may favor some reaction more than other in parallel reactions. 6.True
Answered by See-saw Mind - Fri Feb 6 08:15:27 2009
Chemistry Help Greatly Needed! (Chemical kinetics, rate laws, reaction rates...)?
Q. 1) C + D <---> CD (fast) CD + D ---> CD2 (slow) [ *2 is a subscript] CD2 + D ---> CD3 (fast) [*2 and 3 are subscripts] Based on this mechanism, determine the rate law for the overall reaction. - I know that the slow step's rate law governs the overall rate law, but I do not understand the step involving "Substitute for the intermediates." How do I reach the overall rate law? 2) A + B --> C + D 2C ---> E E + A --->3B + F What is the overall reaction if E is an intermediate?: ~ 4A ---> 2C + D + 3F ~ 4A ---> C + 3D + F ~ A + B ---> C + D + E + F ~ 8A ---> 6D + E + 2F ~ 4A ---> 6D + C + 2F - I know that intermediates don't show in the overall rate law, but I don't know the process of so-called "substitution." 3) If the activation… [cont.]
Asked by lifelectric - Sat Apr 5 01:50:51 2008 - - 1 Answers - 0 Comments
A. Hi there. Well I Think I can answer 2 of the questions. I don't really understand the second one But i'm really tired right now. 1. You need to apply the equilibrium approximation to the first step so you get K1[C][D] = K-1[CD] [CD] = K1[C][D]/K-1 So now you use the steady state approximation on the second step to get d[CD2]/dt = K2[CD][D] - K3 [ CD2][D] = 0 so then [CD2] = K2K1[CD] / K-1K3 at this point you would substitute CD from reaction 1 in and now you have a equation for the intermediate CD2 finally you can find the overall rate law by plugging [CD2] into the equation V = K3[CD2][D] 3. by reverse it means how much energy would it take to get back to reactants from products. The activation energy of the reverse would be the… [cont.]
Answered by Jamie L - Sat Apr 5 02:48:59 2008
Q. 1) C + D <---> CD (fast) CD + D ---> CD2 (slow) [ *2 is a subscript] CD2 + D ---> CD3 (fast) [*2 and 3 are subscripts] Based on this mechanism, determine the rate law for the overall reaction. - I know that the slow step's rate law governs the overall rate law, but I do not understand the step involving "Substitute for the intermediates." How do I reach the overall rate law? 2) A + B --> C + D 2C ---> E E + A --->3B + F What is the overall reaction if E is an intermediate?: ~ 4A ---> 2C + D + 3F ~ 4A ---> C + 3D + F ~ A + B ---> C + D + E + F ~ 8A ---> 6D + E + 2F ~ 4A ---> 6D + C + 2F - I know that intermediates don't show in the overall rate law, but I don't know the process of so-called "substitution." 3) If the activation… [cont.]
Asked by lifelectric - Sat Apr 5 01:50:51 2008 - - 1 Answers - 0 Comments
A. Hi there. Well I Think I can answer 2 of the questions. I don't really understand the second one But i'm really tired right now. 1. You need to apply the equilibrium approximation to the first step so you get K1[C][D] = K-1[CD] [CD] = K1[C][D]/K-1 So now you use the steady state approximation on the second step to get d[CD2]/dt = K2[CD][D] - K3 [ CD2][D] = 0 so then [CD2] = K2K1[CD] / K-1K3 at this point you would substitute CD from reaction 1 in and now you have a equation for the intermediate CD2 finally you can find the overall rate law by plugging [CD2] into the equation V = K3[CD2][D] 3. by reverse it means how much energy would it take to get back to reactants from products. The activation energy of the reverse would be the… [cont.]
Answered by Jamie L - Sat Apr 5 02:48:59 2008
The rate constant for the elementary reaction is 7.1 109/M s at 25 C.?
Q. 2 NO(g) + O2(g) 2 NO2(g) What is the rate constant for the reverse reaction at the same temperature?
Asked by Kyle R - Mon Jul 14 23:42:31 2008 - - 1 Answers - 0 Comments
A. This reaction is known to have an equilibrium constant Kc at 298 K (25 C) of 4.67x10^13 (see source below). For reversible reactions, Keq = k(forward) / k(reverse); 4.67x10^13 = 7.1x10^9 / k(reverse); k(reverse) = 7000.0 /M s.
Answered by Glenguin - Thu Jul 17 22:32:51 2008
Q. 2 NO(g) + O2(g) 2 NO2(g) What is the rate constant for the reverse reaction at the same temperature?
Asked by Kyle R - Mon Jul 14 23:42:31 2008 - - 1 Answers - 0 Comments
A. This reaction is known to have an equilibrium constant Kc at 298 K (25 C) of 4.67x10^13 (see source below). For reversible reactions, Keq = k(forward) / k(reverse); 4.67x10^13 = 7.1x10^9 / k(reverse); k(reverse) = 7000.0 /M s.
Answered by Glenguin - Thu Jul 17 22:32:51 2008
The chemical reaction N2(g) + 3H2(g) <--> 2NH3(g) is at equilibrium in a reaction chamber.?
Q. If an experimenter injects a small amount of N2 into the reaction chamber, what happens to both the forward and reverse reaction rates from the moment the N2 is added until the equilibrium is re-established.
Asked by snaredrum_girl - Thu May 29 00:08:10 2008 - - 1 Answers - 0 Comments
Q. If an experimenter injects a small amount of N2 into the reaction chamber, what happens to both the forward and reverse reaction rates from the moment the N2 is added until the equilibrium is re-established.
Asked by snaredrum_girl - Thu May 29 00:08:10 2008 - - 1 Answers - 0 Comments
What is the activation energy for the forward and reverse reactions?
Q. At 650K, rate constant for forward reaction HI(g)=1/2H2 (g)+ 1/2 I2(g) is k=2.15 x 10^-8 L/mol sec. At 700K, rate constant is 2.39 x 10^-7 1/Ms the delta H for reverse is 5.12 kJ/mol for reverse rxn.
Asked by Mickey - Thu Aug 14 16:18:45 2008 - - 1 Answers - 0 Comments
A. Use Arrhenius equation to determine activation energy of forward reaction: k (T ) = A exp{- Ea /(R T )} k (T ) = A exp{- Ea /(R T )} => k (T )/k (T ) = A exp{- Ea /(R T )} / A exp{- Ea/(R T } <=> k (T )/k (T ) = exp{ (Ea /R) (1/T - 1/T ) } <=> ln(k (T )/k (T )) = (Ea /R) (1/T - 1/T ) => Ea = R ln(k (T )/k (T )) / (1/T - 1/T ) = 8.314472Jmol K ln(2.39 10 Lmol s / 2.15 10 Lmol s ) / ( 1/650K - 1/750K) = 182.22 kJ/mol The difference of the activation energy is equal to th enthalpy of reaction. You can deduce this by considering the equilibrium constant of the reaction. By kinetic definition it is defined as the ratio of forward to reverse rate constant: K = k / k = A exp{- Ea /(R T)} / A exp{- Ea /(R T)} = (A /A ) exp{- (Ea -… [cont.]
Answered by schmiso - Mon Aug 18 07:56:21 2008
Q. At 650K, rate constant for forward reaction HI(g)=1/2H2 (g)+ 1/2 I2(g) is k=2.15 x 10^-8 L/mol sec. At 700K, rate constant is 2.39 x 10^-7 1/Ms the delta H for reverse is 5.12 kJ/mol for reverse rxn.
Asked by Mickey - Thu Aug 14 16:18:45 2008 - - 1 Answers - 0 Comments
A. Use Arrhenius equation to determine activation energy of forward reaction: k (T ) = A exp{- Ea /(R T )} k (T ) = A exp{- Ea /(R T )} => k (T )/k (T ) = A exp{- Ea /(R T )} / A exp{- Ea/(R T } <=> k (T )/k (T ) = exp{ (Ea /R) (1/T - 1/T ) } <=> ln(k (T )/k (T )) = (Ea /R) (1/T - 1/T ) => Ea = R ln(k (T )/k (T )) / (1/T - 1/T ) = 8.314472Jmol K ln(2.39 10 Lmol s / 2.15 10 Lmol s ) / ( 1/650K - 1/750K) = 182.22 kJ/mol The difference of the activation energy is equal to th enthalpy of reaction. You can deduce this by considering the equilibrium constant of the reaction. By kinetic definition it is defined as the ratio of forward to reverse rate constant: K = k / k = A exp{- Ea /(R T)} / A exp{- Ea /(R T)} = (A /A ) exp{- (Ea -… [cont.]
Answered by schmiso - Mon Aug 18 07:56:21 2008
Colliding molecules with the capability to cause a reaction to occur.?
Q. 1. Colliding molecules with the capability to cause a reaction to occur. a. activation energy b. catalyst c. effective collision d. endergonic e. entropy f. exergonic g. inhibitor h. position of equilibrium i. reaction mechanism j. reaction rate k. spontaneous process l. state of equilibrium 2. A detailed explanation of how a reaction occurs. a. activation energy b. catalyst c. effective collision d. endergonic e. entropy f. exergonic g. inhibitor h. position of equilibrium i. reaction mechanism j. reaction rate k. spontaneous process l. state of equilibrium 3. A process that gives up energy as it takes place. a. activation energy b. catalyst c. effective collision d. endergonic e. entropy f. exergonic g. inhibitor h. position of… [cont.]
Asked by CrystalSuga - Wed Apr 1 10:40:57 2009 - - 1 Answers - 0 Comments
A. 1. effective collision 2.reaction mechanism 3.exergonic 4.state of equilibrium
Answered by science teacher - Wed Apr 1 10:47:00 2009
Q. 1. Colliding molecules with the capability to cause a reaction to occur. a. activation energy b. catalyst c. effective collision d. endergonic e. entropy f. exergonic g. inhibitor h. position of equilibrium i. reaction mechanism j. reaction rate k. spontaneous process l. state of equilibrium 2. A detailed explanation of how a reaction occurs. a. activation energy b. catalyst c. effective collision d. endergonic e. entropy f. exergonic g. inhibitor h. position of equilibrium i. reaction mechanism j. reaction rate k. spontaneous process l. state of equilibrium 3. A process that gives up energy as it takes place. a. activation energy b. catalyst c. effective collision d. endergonic e. entropy f. exergonic g. inhibitor h. position of… [cont.]
Asked by CrystalSuga - Wed Apr 1 10:40:57 2009 - - 1 Answers - 0 Comments
A. 1. effective collision 2.reaction mechanism 3.exergonic 4.state of equilibrium
Answered by science teacher - Wed Apr 1 10:47:00 2009
Which of the factors below is not a condition encessary for equilibrium?
Q. a) a closed system b) a constant temperature c) equal forward and reverse reaction rates d) equal concentrations of reactants and products and why?
Asked by asbquestion02 - Mon Nov 6 22:56:05 2006 - - 2 Answers - 0 Comments
A. I think you are in my class. -0-... do you have chem test tommorow? I believe it is d. Equilibrium has to be in a closed system. Temperature changes K value so it is necessary to keep in a constant temperature. Equilibrium basically is defined as equal forward and reverse reaction reates. Concentration does not necessary has to be same.
Answered by whatzzup012 - Mon Nov 6 23:00:15 2006
Q. a) a closed system b) a constant temperature c) equal forward and reverse reaction rates d) equal concentrations of reactants and products and why?
Asked by asbquestion02 - Mon Nov 6 22:56:05 2006 - - 2 Answers - 0 Comments
A. I think you are in my class. -0-... do you have chem test tommorow? I believe it is d. Equilibrium has to be in a closed system. Temperature changes K value so it is necessary to keep in a constant temperature. Equilibrium basically is defined as equal forward and reverse reaction reates. Concentration does not necessary has to be same.
Answered by whatzzup012 - Mon Nov 6 23:00:15 2006
How do the rates of the forward and reverse reactions compare at a state of dynamic chemical equilibrium?
Q. How do the rates of the forward and reverse reactions compare at a state of dynamic chemical equilibrium?
Asked by munchkin - Sat Aug 4 01:23:06 2007 - - 4 Answers - 0 Comments
A. They are equal - that's the definition of equilibrium
Answered by KennyB - Sat Aug 4 01:30:43 2007
Q. How do the rates of the forward and reverse reactions compare at a state of dynamic chemical equilibrium?
Asked by munchkin - Sat Aug 4 01:23:06 2007 - - 4 Answers - 0 Comments
A. They are equal - that's the definition of equilibrium
Answered by KennyB - Sat Aug 4 01:30:43 2007
Which of the following is not a mechanism for an enzyme catalyzing the reaction A+B->C?
Q. a.The substrates, A and B, bind the active site and are brought in close proximity of each other b.The substrates, A and B, make temporary covalent bonds with amino acids in the active site of the protein c.The substrates, A and B, bind the active site and effectively block the reverse reaction d.The substrates A and B bind the active site and are exposed to a different local environment which increases the rate of the reaction
Asked by TJ - Wed Oct 7 16:17:37 2009 - - 1 Answers - 0 Comments
A. b. The substrates, A and B, make temporary covalent bonds with amino acids in the active site of the protein
Answered by Peter S - Fri Oct 9 06:53:52 2009
Q. a.The substrates, A and B, bind the active site and are brought in close proximity of each other b.The substrates, A and B, make temporary covalent bonds with amino acids in the active site of the protein c.The substrates, A and B, bind the active site and effectively block the reverse reaction d.The substrates A and B bind the active site and are exposed to a different local environment which increases the rate of the reaction
Asked by TJ - Wed Oct 7 16:17:37 2009 - - 1 Answers - 0 Comments
A. b. The substrates, A and B, make temporary covalent bonds with amino acids in the active site of the protein
Answered by Peter S - Fri Oct 9 06:53:52 2009
Which of the following is correct?(10 ptz and 5 stars for explaining your answer>i promice?
Q. An equilibrium system shifts left when the... a. rate of the forward reaction is equal to the rate of the reverse reaction. b. rate of the forward reaction is less than the rate of the reverse reaction. c. rate of the forward reaction is greater than the rate of the reverse reaction. d. rate of the forward reaction and the rate of the reverse reaction are constant.
Asked by Elisabeth W - Sat Nov 1 13:46:17 2008 - - 1 Answers - 0 Comments
A. a. Because Le Chatelier's principle allows us to predict the effects of changes in temperature, pressure, and concentration on a system at equilibrium. It states that if a system at equilibrium experiences a change, the system will shift its equilibrium to try to compensate for the change. PS:the reverse reaction will happen the most
Answered by antijerryforce1 - Sat Nov 1 14:37:44 2008
Q. An equilibrium system shifts left when the... a. rate of the forward reaction is equal to the rate of the reverse reaction. b. rate of the forward reaction is less than the rate of the reverse reaction. c. rate of the forward reaction is greater than the rate of the reverse reaction. d. rate of the forward reaction and the rate of the reverse reaction are constant.
Asked by Elisabeth W - Sat Nov 1 13:46:17 2008 - - 1 Answers - 0 Comments
A. a. Because Le Chatelier's principle allows us to predict the effects of changes in temperature, pressure, and concentration on a system at equilibrium. It states that if a system at equilibrium experiences a change, the system will shift its equilibrium to try to compensate for the change. PS:the reverse reaction will happen the most
Answered by antijerryforce1 - Sat Nov 1 14:37:44 2008
reaction changes?
Q. At chemical equilibrium: a)the product concentration equals the reactant concentration b)the rate of the forward reaction equals the rate of the reverse reaction c)neither the forward reaction or the reverse reaction is occurring at all A reaction that takes place in 1 ms compared to a reaction that takes place in a year: a)will have a higher Keq b)will have a lower Keq c)will have a higher rate How does the concentration of a pure liquid change throughout the course of a reaction? a)it depends on the number of moles in the reactant vs. the number of moles in the product b)it decreases because the substance is no longer pure c)it does not change
Asked by Vic - Sat Jun 9 02:23:57 2007 - - 2 Answers - 0 Comments
A. at equilibrium forward rate is equals to the backward rate thats why it is equilibrium first one is b) it has a higher reaction rate we dont even know that it is a reversible reaction also keq is not a measure of how fast a reaction takes place so answer is c) reaction means producing other species so it wont be pure anymore and concentration will decrease b)
Answered by shaq - Sat Jun 9 04:51:20 2007
Q. At chemical equilibrium: a)the product concentration equals the reactant concentration b)the rate of the forward reaction equals the rate of the reverse reaction c)neither the forward reaction or the reverse reaction is occurring at all A reaction that takes place in 1 ms compared to a reaction that takes place in a year: a)will have a higher Keq b)will have a lower Keq c)will have a higher rate How does the concentration of a pure liquid change throughout the course of a reaction? a)it depends on the number of moles in the reactant vs. the number of moles in the product b)it decreases because the substance is no longer pure c)it does not change
Asked by Vic - Sat Jun 9 02:23:57 2007 - - 2 Answers - 0 Comments
A. at equilibrium forward rate is equals to the backward rate thats why it is equilibrium first one is b) it has a higher reaction rate we dont even know that it is a reversible reaction also keq is not a measure of how fast a reaction takes place so answer is c) reaction means producing other species so it wont be pure anymore and concentration will decrease b)
Answered by shaq - Sat Jun 9 04:51:20 2007
A question about equlibrium?
Q. Can you please explain the following: 2SO2(g)+O2(l)<-->2SO3(g)+ energy Rate theory explains the result of cooling this exothermic system by assuming that both forward and reverse reaction rates are slower at the lower temperature, but that the reverse rate decreases more than the forward rate. While the rates remain unequal, the observed result is the production of more product and more energy. The shift causes concentration changes that will increase the reverse rate and decrease the forward rate until they become equal again, at the new, lower temperature.
Asked by Kate - Sat Jan 20 20:31:17 2007 - - 1 Answers - 0 Comments
A. The equalibrium must be preserved. Since you are lowering the temperature (effectively pushing/forcing the reaction to move in the reverse direction) the sytstem will respond by increasing the concentration in SO3 until equalibium is re-established. I hope that explains it.
Answered by Dr Dave P - Sat Jan 20 20:43:35 2007
Q. Can you please explain the following: 2SO2(g)+O2(l)<-->2SO3(g)+ energy Rate theory explains the result of cooling this exothermic system by assuming that both forward and reverse reaction rates are slower at the lower temperature, but that the reverse rate decreases more than the forward rate. While the rates remain unequal, the observed result is the production of more product and more energy. The shift causes concentration changes that will increase the reverse rate and decrease the forward rate until they become equal again, at the new, lower temperature.
Asked by Kate - Sat Jan 20 20:31:17 2007 - - 1 Answers - 0 Comments
A. The equalibrium must be preserved. Since you are lowering the temperature (effectively pushing/forcing the reaction to move in the reverse direction) the sytstem will respond by increasing the concentration in SO3 until equalibium is re-established. I hope that explains it.
Answered by Dr Dave P - Sat Jan 20 20:43:35 2007
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