Question

Consider the reactionA + B → ProductsFrom the following data obtained at a certain temperature, determine the order of the reaction. Enter the order with respect to A, the order with respect to B, and the overall reaction order.[A] (M) [B] (M) Rate (M/s)1.50 1.50 3.20 ×10−11.50 2.50 3.20 ×10−13.00 1.50 6.40 ×10−1

Answer 1

Answer: Order with respect to A is 1 , order with respect to B is 0 and total order is 1

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

$Rate=k[A]^x[B]^y$

k= rate constant

x = order with respect to A

y = order with respect to A

n = x+y = Total order

a) From trial 1: $3.20\times 10^{-1}=k[1.50]^x[1.50]^y$    (1)

From trial 2: $3.20\times 10^{-1}=k[1.50]^x[2.50]^y$    (2)

Dividing 2 by 1 :$\frac{3.20\times 10^{-1}}{3.20\times 10^{-1}}=\frac{k[1.50]^x[2.50]^y}{k[1.50]^x[1.50]^y}$

$1=1.66^y,2^0=1.66^y$ therefore y=0

b) From trial 2: $3.20\times 10^{-1}=k[1.50]^x[2.50]^y$      (3)

From trial 3: $6.40\times 10^{-1}=k[3.00]^x[1.50]^y$   (4)

Dividing 4 by 3:$\frac{6.40\times 10^{-1}}{3.20\times 10^{-1}}=\frac{k[3.00]^x[1.50]^y}{k[1.50]^x[2.50]^y}$

$2=2^x,2^1=2^x$, x=1

Thus rate law is $Rate=k[A]^1[B]^0$

Thus order with respect to A is 1 , order with respect to B is 0 and total order is 1+0=1.