Question 12.5: Using the Integrated Rate Law for a First-Order Reaction The......
Using the Integrated Rate Law for a First-Order Reaction
The decomposition of hydrogen peroxide in dilute sodium hydroxide solution is described by the equation
2 H_2O_2(aq) → 2 H_2O(l) + O_2(g)
The reaction is first order in H_2O_2, the rate constant for the consumption of H_2O_2 at 20 °C is 1.8 × 10^{-5} \ s^{-1} , and the initial concentration of H_2O_2 is 0.30 M.
(a) What is the concentration of H_2O_2 after 4.00 h?
(b) How long will it take for the H_2O_2 concentration to drop to 0.12 M?
(c) How long will it take for 90% of the H_2O_2 to decompose?
STRATEGY
Since this reaction has a first-order rate law, -Δ[ H_2O_2 ]/Δt = k[ H_2O_2 ], we can use the corresponding concentration–time equation for a first-order reaction:
\ln \frac{\left[\mathrm{H}_2 \mathrm{O}_2\right]_t}{\left[\mathrm{H}_2 \mathrm{O}_2\right]_0}=-k tIn each part, we substitute the known quantities into this equation and solve for the unknown.
(a) Because k has units of s^{-1} , we must first convert the time from hours to seconds:
t=(4.00 \mathrm{~h})\left(\frac{60 \mathrm{~min}}{\mathrm{~h}}\right)\left(\frac{60 \mathrm{~s}}{\min }\right)=1.44 \times 10^4 \mathrm{~s}Ten, substitute the values of [ H_2O_2]_0, k, and t into the concentration–time equation:
\ln \frac{\left[\mathrm{H}_2 \mathrm{O}_2\right]_t}{0.30 \mathrm{M}}=-\left(1.8 \times 10^{-5} \mathrm{~s}^{-1}\right)\left(1.44 \times 10^4 \mathrm{~s}\right)=-0.259Taking the natural antilogarithm (antiln) of both sides gives
\frac{\left[\mathrm{H}_2 \mathrm{O}_2\right]_t}{0.30 \mathrm{M}}=e^{-0.259}=0.772where the number e = 2.718 28 . . . the base of natural logarithms (see Appendix A.2). Therefore,
\left[\mathrm{H}_2 \mathrm{O}_2\right]_t=(0.772)(0.30 \mathrm{M})=0.23 \mathrm{M}(b) First, solve the concentration–time equation for the time:
t=-\frac{1}{k} \ln \frac{\left[\mathrm{H}_2 \mathrm{O}_2\right]_t}{\left[\mathrm{H}_2 \mathrm{O}_2\right]_0}Ten evaluate the time by substituting the concentrations and the value of k:
t=-\left(\frac{1}{1.8 \times 10^{-5} \mathrm{~s}^{-1}}\right)\left(\ln \frac{0.12 \ \mathrm{M}}{0.30 \ \mathrm{M}}\right)=-\left(\frac{1}{1.8 \times 10^{-5} \mathrm{~s}^{-1}}\right)(-0.916)=5.1 \times 10^4 \mathrm{~s}Thus, the H_2O_2 concentration reaches 0.12 M at a time of 5.1 × 10^4 s (14 h).
(c) When 90% of the H_2O_2 has decomposed, 10% remains. Therefore,
\frac{\left[\mathrm{H}_2 \mathrm{O}_2\right]_t}{\left[\mathrm{H}_2 \mathrm{O}_2\right]_0}=\frac{(0.10)(0.30 \mathrm{M})}{(0.30 \mathrm{M})}=0.10Te time required for 90% decomposition is
t=-\left(\frac{1}{1.8 \times 10^{-5} \mathrm{~s}^{-1}}\right)(\ln 0.10)=-\left(\frac{1}{1.8 \times 10^{-5} \mathrm{~s}^{-1}}\right)(-2.30)=1.3 \times 10^5 \mathrm{~s}(36 \mathrm{~h})BALLPARK CHECK
The concentration of H_2O_2 (0.23 M) after 4.00 h is less than the initial concentration (0.30 M). A longer period of time (14 h) is required for the concentration to drop to 0.12 M, and still more time (36 h) is needed for the concentration to fall to 0.030 M (10% of the original concentration). These results look reasonable. A plot of [ H_2O_2 ] versus time would exhibit an exponential decay of the H_2O_2 concentration, as expected for a first order reaction.