Question

Determine the rate law and the value of k for the following reaction using the data provided.CO(g) Cl2(g) → COCl2(g)[CO]i (M)[Cl2]i(M)Initial Rate (M-1s-1)0.250.400.6960.250.801.970.500.803.94A) Rate

Answer 1

Answer: The rate law expression is $\text{Rate}=k[CO]^1[Cl_2]^{\frac{3}{2}}$ and value of 'k' is $11.04M^{-\frac{7}{2}}s^{-1}$

Explanation:

Rate law 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:

$CO(g)+Cl_2(g)\rightarrow COCl_2(g)$

Rate law expression for the reaction:

$\text{Rate}=k[CO]^a[Cl_2]^b$

where,

a = order with respect to carbon monoxide

b = order with respect to chlorine

• Expression for rate law for first observation:

$0.696=k(0.25)^a(0.40)^b$       ....(1)

• Expression for rate law for second observation:

$1.970=k(0.25)^a(0.80)^b$       ....(2)

• Expression for rate law for third observation:

$3.94=k(0.50)^a(0.80)^b$       ....(3)

Dividing 2 from 3, we get:

$\frac{3.94}{1.970}=\frac{(0.50)^a(0.80)^b}{(0.50)^a(0.80)^b}\\\\2=2^a\\a=1$

Dividing 1 from 2, we get:

$\frac{1.970}{0.696}=\frac{(0.25)^a(0.80)^b}{(0.25)^a(0.40)^b}\\\\2.83=2^b\\b=1.5$

Thus, the rate law becomes:

$\text{Rate}=k[CO]^1[Cl_2]^{\frac{3}{2}}$

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

Putting values in equation 1, we get:

$0.696=k[0.25]^1[0.40]^{\frac{3}{2}}\\\\k=11.04M^{-\frac{7}{2}}s^{-1}$

Hence, the rate law expression is $\text{Rate}=k[CO]^1[Cl_2]^{\frac{3}{2}}$ and value of 'k' is $11.04M^{-\frac{7}{2}}s^{-1}$