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A student performed a chemical reaction which involved 3 reactants: A, B and C and measured...

A student performed a chemical reaction which involved 3 reactants: A, B and C and measured the initial reaction rate but varied initial concentrations. The results showed that doubling the concentration of B doubled the reaction rate, doubling the concentration of C quadrupled the reaction rate and doubling the concentration of A had no effect on the rate of reaction.

Find the overall reaction order and write the rate equation.

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Answer #1

Consider given reaction, A + B + C \rightarrow Product

Rate law expression for above reaction is Rate of reaction = k [ A] x [ B] y [ C] z , where k is rate constant of reaction.

From above expression , we can write Rate 1 = k [ A] x [ B] y [ C] z

The concentration of B is doubled , so that it becomes 2 [ B ].

The new rate law is , Rate 2 = k [ A] x ( 2 [ B] ) y [ C] z

\therefore Rate 2 / Rate 1 = k [ A] x ( 2 [ B] ) y [ C] z / k [ A] x [ B] y [ C] z

Rate 2 / Rate 1 = 2 y k [ A] x [ B] y [ C] z / k [ A] x [ B] y [ C] z

Rate 2 / Rate 1 = 2 y .

On doubling concentration of B , rate of reaction doubles. Therefore, we get  Rate 2 / Rate 1 = 2 y = 2

Therefore, y = 1

i e order of reaction w.r.t B is 1.

The concentration of C is doubled , so that it becomes 2 [ C ].

The new rate law is , Rate 3 = k [ A] x  [ B] y ( 2 [ C] ) z

Hence, Rate 3 / Rate 1 = k [ A] x  [ B] y ( 2 [ C] ) z / k [ A] x [ B] y [ C] z

Rate 3 / Rate 1 = 2 z k [ A] x [ B] y [ C] z / k [ A] x [ B] y [ C] z

Rate 3 / Rate 1 = 2 z .

On doubling concentration of C , rate of reaction quadruples. Therefore, we get  Rate 3 / Rate 1 = 2 y = 4

Therefore, y =2

i e order of reaction w.r.t C is 2.

Doubling concentration of C have no effect on rate of reaction means order of reaction w.r.t A is zero.

From x,y,z values we get rate law expression for given reaction as Rate = k [B] [ C] 2

Overall order of reaction is 3.

ANSWER : Rate = k [B] [ C] 2 & Overall order of reaction= 3.

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