Question

The following data were obtained for the reaction 2A + B → C where rate = Δ[C]/Δt [A](M) [B](M) Initial Rate (M/s) 0.100 0.0500 2.13 × 10–4 0.200 0.0500 1.70 × 10–2 0.400 0.100 1.36 × 10–1 What is the value of the rate constant?

Answer 1

Answer :  The value of the rate constant 'k' for this reaction is $2.67\times 10^{2}M^{-2}s^{-1}$

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:

$2A+B\rightarrow C$

Rate law expression for the reaction:

$\text{Rate}=k[A]^a[B]^b$

where,

a = order with respect to A

b = order with respect to B

Expression for rate law for first observation:

$2.13\times 10^{-4}=k(0.100)^a(0.0500)^b$ ....(1)

Expression for rate law for second observation:

$1.70\times 10^{-2}=k(0.200)^a(0.0500)^b$ ....(2)

Expression for rate law for third observation:

$1.36\times 10^{-1}=k(0.400)^a(0.100)^b$ ....(3)

Dividing 2 by 1, we get:

$\frac{1.70\times 10^{-2}}{2.13\times 10^{-4}}=\frac{k(0.200)^a(0.0500)^b}{k(0.100)^a(0.0500)^b}\\\\80=2^a\\\\\log 80=a\times \log 2\\\\a=6$

Dividing 3 by 2 and put value of a = 6, we get:

$\frac{1.36\times 10^{-1}}{1.70\times 10^{-2}}=\frac{k(0.400)^6(0.100)^b}{k(0.200)^6(0.0500)^b}\\\\8=2^{(b+6)}\\\\b+6=3\\\\b=-3$

Thus, the rate law becomes:

$\text{Rate}=k[A]^a[B]^b$

$\text{Rate}=k[A]^6[B]^{-3}$

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

$2.13\times 10^{-4}=k(0.100)^6(0.0500)^{-3}$

$k=2.67\times 10^{2}M^{-2}s^{-1}$

Hence, the value of the rate constant 'k' for this reaction is $2.67\times 10^{2}M^{-2}s^{-1}$