Question

Which one of the following is a rate equation for second order bimolecular reaction if, a and b are the initial concentrations of A and B, respectively, and x is the concentration of each species reacting in time t and k is second‐order reaction

• A. k = [2.303/ t(a − b)] x [log a(a − x) / b(b − x) ]
• B. k = [2.303/ t(a − b)] x [log b(a − x) / a(b − x) ]
• C. k = [2.303/ t(a − b)] / [log b(a − x) / a(b − x) ]
• D. k = [2.303/ t(a − b)] / [log a(a − x) / b(b − x) ]

Answer 1

The Correct Option is C

To determine the correct rate equation for a second-order bimolecular reaction involving the reactants A and B, we need to evaluate the provided options. The equations provided express the rate constant \(k\) in terms of the initial concentrations of A and B, the amount reacted \(x\) at time \(t\), and the differences in these concentrations. 

The rate equation for a second-order reaction with two different reactants, A and B, can be given by:

\(\displaystyle \frac{1}{a - b} \ln \left( \frac{b(a-x)}{a(b-x)} \right) = kt\)

Rearranging for \(k\) gives us:

\(k = \frac{1}{t(a-b)} \ln \left( \frac{b(a-x)}{a(b-x)} \right)\)

To transform this into a suitable format with base 10 logarithms, we use:

\(\ln(x) = 2.303 \log_{10}(x)\)

Thus, the equation becomes:

\(k = \frac{2.303}{t(a-b)} \log \left( \frac{b(a-x)}{a(b-x)} \right)\)

Let's evaluate the given options to find the match:

\(k = \frac{2.303}{t(a-b)} \log \left( \frac{b(a-x)}{a(b-x)} \right)\)

• Option 1: \(k = \frac{2.303}{t(a-b)} \log \left( \frac{a(a-x)}{b(b-x)} \right)\) - Incorrect structure.
• Option 2: \(k = \frac{2.303}{t(a-b)} \log \left( \frac{b(a-x)}{a(b-x)} \right)\) - Correct structure.
• Option 3 and Option 4: both incorrect as they involve division by log rather than multiplication.

Thus, Option 2, \(k = \frac{2.303}{t(a-b)} \log \left( \frac{b(a-x)}{a(b-x)} \right)\), is the correct answer. This option correctly rearranges the rate equation in the form suitable for base 10 logarithms.