Question

For the reaction\({\bf{Q}} \to {\bf{W + X}}\), the following data were obtained at 30 °C

1. What is the order of the reaction with respect to (Q), and what is the rate law?
2. What is the rate constant?

Short answer

1. The order of reaction is 2 and the rate law is represented as \({\bf{rate = }}0.23{{\bf{(Q)}}^{\bf{2}}}\).
2. value of rate constant is\({\bf{\;0}}{\bf{.230mol}}{{\bf{L}}^{{\bf{ - 1}}}}{\bf{mi}}{{\bf{n}}^{{\bf{ - 1}}}}\)

Step-by-step solution

Rate law

The rate law for a chemical reaction is an expression that provides a relationship between the rate of the reaction and the concentration of the reactants participating in it.

General rate law for the given reaction is represented as

\({\bf{rate = }}k{{\bf{(Q)}}^{\bf{n}}}\)

From the given table we can make the following equations

\(\begin{align}6.68 \times {10^{ - 3}}mol{L^{ - 1}}{s^{ - 1}} &= k{(0.170)^n}\,\,\,\,\,\,......(1)\\1.04 \times {10^{ - 2}}mol{L^{ - 1}}{s^{ - 1}} &= k{(0.212)^n}\,\,\,\,\,\,\,......(2)\\2.94 \times {10^{ - 2}}mol{L^{ - 1}}{s^{ - 1}} &= k{(0.357)^n}\,\,\,\,\,\,\,......(3)\end{align}\)

Order of reaction

The order of reaction refers to the power dependence of the rate on the concentration of each reactant.

Order of reaction can be calculated as:

\(\begin{align}1.04 \times {10^{ - 2}}mol{L^{ - 1}}{s^{ - 1}} &= K{(0.212M)^n}\\E\lim inate\,\,k\\6.68 \times {10^{ - 3}} &= \frac{{1.04 \times {{10}^{ - 2}}mol{L^{ - 1}}{s^{ - 1}}}}{{{{(0.212M)}^n}}}{(0.170)^n}\end{align}\)

\(\begin{align}0.6423 &= {(0.8019)^n}\\taking\,\,natural\,\log \\ - 0.4427 &= n( - 0.2208)\\n &= 2\end{align}\)

Hence, the reaction is of second order.

Rate constant

The rate constant is the proportionality constant in rate equation.

Rate constant can be calculated as:

\(\begin{align}6.68 \times {10^{ - 3}}mol{L^{ - 1}}{s^{ - 1}} &= K{(1.70M)^2}\\K &= \;0.230mol{L^{ - 1}}{\min ^{ - 1}}\end{align}\)