Question

The reaction, 2 NO(g) + Cl2(g) → 2 NOCl(g) was studied at -10oC. The following results were obtained. [NO] (M) [Cl2] (M) Initial Rate (M min-1) 0.10 0.10 0.18 0.10 0.20 0.36 0.20 0.20 1.45 a) What is the rate law? b) What is the value of the rate constant? (include correct units)

Answer 1

Answer :

(a) The rate law becomes:

$\text{Rate}=k[NO]^2[Cl_2]^1$

(b) The value of the rate constant 'k' for this reaction is $180M^{-2}min^{-1}$

Explanation :

Rate law : It is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

$2NO(g)+Cl_2(g)\rightarrow 2NOCl(g)$

Rate law expression for the reaction:

$\text{Rate}=k[NO]^a[Cl_2]^b$

where,

a = order with respect to $NO$

b = order with respect to $Cl_2$

Expression for rate law for first observation:

$0.18=k(0.10)^a(0.10)^b$ ....(1)

Expression for rate law for second observation:

$0.36=k(0.10)^a(0.20)^b$ ....(2)

Expression for rate law for third observation:

$1.45=k(0.20)^a(0.20)^b$ ....(3)

Dividing 1 from 2, we get:

$\frac{0.36}{0.18}=\frac{k(0.10)^a(0.20)^b}{k(0.10)^a(0.10)^b}\\\\2=2^b\\b=1$

Dividing 2 from 3, we get:

$\frac{1.45}{0.36}=\frac{k(0.20)^a(0.20)^b}{k(0.10)^a(0.20)^b}\\\\4=2^a\\a=2$

(a) Thus, the rate law becomes:

$\text{Rate}=k[NO]^2[Cl_2]^1$

(b) Now, calculating the value of 'k' by using any expression.

Putting values in equation 1, we get:

$0.18=k(0.10)^2(0.10)^1$

$k=180M^{-2}min^{-1}$

Hence, the value of the rate constant 'k' for this reaction is $180M^{-2}min^{-1}$