Question

At $ T $ (K), the following data was obtained for the reaction: $ S_2O_8^{2-} + 3 I^- \rightarrow 2 SO_4^{2-} + I_3^- $.

From the data, the rate constant of the reaction (in $ M^{-1} s^{-1} $) is: 

• A. \(8.0 \times 10^{-2}\)
• B. \(4.0 \times 10^{-2}\)
• C. \(8.0 \times 10^{-3}\)
• D. \(6.0 \times 10^{-3}\)

Hint:

To determine the rate constant, first determine the reaction order by comparing experimental data, then use the rate law equation to calculate \( k \).

Answer 1

The Correct Option is A

The rate law for the reaction is assumed to be: \[ {Rate} = k [S_2O_8^{2-}]^m [I^-]^n \] Step 1: Determine the order of reaction with respect to \( I^- \) Comparing Experiments 1 and 2: \[ \frac{{Rate}_1}{{Rate}_2} = \left(\frac{[I^-]_1}{[I^-]_2}\right)^n \] \[ \frac{2.2 \times 10^{-4}}{1.1 \times 10^{-4}} = \left(\frac{0.034}{0.017}\right)^n \] \[ 2 = 2^n \Rightarrow n = 1 \] Step 2: Determine the order of reaction with respect to \( S_2O_8^{2-} \) Comparing Experiments 2 and 3: \[ \frac{{Rate}_3}{{Rate}_2} = \left(\frac{[S_2O_8^{2-}]_3}{[S_2O_8^{2-}]_2}\right)^m \] \[ \frac{2.2 \times 10^{-4}}{1.1 \times 10^{-4}} = \left(\frac{0.160}{0.080}\right)^m \] \[ 2 = 2^m \Rightarrow m = 1 \] Thus, the rate law is: \[ {Rate} = k [S_2O_8^{2-}] [I^-] \] Step 3: Calculate \( k \) Using Experiment 1: \[ 2.2 \times 10^{-4} = k (0.080) (0.034) \] \[ k = \frac{2.2 \times 10^{-4}}{(0.080 \times 0.034)} \] \[ k = \frac{2.2 \times 10^{-4}}{2.72 \times 10^{-3}} \] \[ k = 8.0 \times 10^{-2} { M}^{-1} { s}^{-1} \] Thus, the correct rate constant is \(8.0 \times 10^{-2}\) M\(^{-1}\) s\(^{-1}\).