1)      Many processes contribute to the BZ reaction:  three are 2H^+ (aq)  + BrO2^-(aq) +  BrO3^- (aq)   <-->    2BrO2(aq) +H2O(I)2Ce^3+ + 2BrO2(aq) <--> 2Ce^4+(aq) +...

1)      Many processes contribute to the BZ reaction:  three are

2H^+ (aq)  + BrO2^-(aq) +  BrO3^- (aq)   <-->    2BrO2(aq) +H2O(I)

2Ce^3+ + 2BrO2(aq) <--> 2Ce^4+(aq) + 2BrO2^-(aq)

2BrO2^-(aq) <---> BrO3^-(aq)+ BrO^-(aq)

a) When these reactions are isolated from the other processes in the BZ reaction, they can go to equilibrium. Write the expression for Kc for each chemical equation for each step as written above

b) Write the overall chemical equation for the three steps, and write K for the overall reaction.

c) What is the relation between the three Kcs of the individual reaction steps and Kc for the overall reaction.

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mlsiasebs | College Teacher | (Level 1) Associate Educator

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a) We can write the Kc expressions for each of the reactions by using the law of mass action - products over reactants, concentrations raised to the power of the coefficients.

For the first reaction, we get (remember, water is not included because it's a pure liquid)

Kc = [BrO2]^2 / [H+]^2 [BrO2-] [BrO3-]

For the second reaction, we will have

Kc = [Ce4+]^2 [BrO2-]^2  /  [Ce3+]^2 [BrO2]^2

And for the third reaction, we will have

Kc = [BrO3-][BrO-] / [BrO2-]^2

b) For the overall chemical reaction, we add up the three reactions and get the following for Kc for the reaction

2H^+ (aq) + BrO2^-(aq) + BrO3^- (aq) <--> 2BrO2(aq) +H2O(l)

2Ce^3+ + 2BrO2(aq) <--> 2Ce^4+(aq) + 2BrO2^-(aq)

2BrO2^-(aq) <---> BrO3^-(aq)+ BrO^-(aq)

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2H^+ (aq) + BrO2^-(aq) + 2Ce^  <--> +H2O(l) + 2Ce^4+(aq) +  BrO^-(aq)

 Kc = [Ce4+]^2   [BrO-] / [Ce3+]^2 [H+]^2 [BrO2-] 

 c) When we add reactions together, we can take the products of the individual Kc values and find the Kc of the overall reaction.  If you look at the product of the three Kc expressions in part A and multiply them, cancelling out like terms, you will end up with the Kc shown in part C.

 

 

 

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