Question 12.5: The decomposition of hydrogen peroxide in dilute sodium hydr......
The decomposition of hydrogen peroxide in dilute sodium hydroxide solution is described by the equation
2 H_{2}O_{2} (aq) \longrightarrow 2 H_{2}O (l) + O_{2} (g)The reaction is first order in H_{2}O_{2} , the rate constant for consumption of H_{2}O_{2} \text{at} 20°C \text{is} 1.8 × 10^{-5} , and the initial concentration of H_{2}O_{2} is 0.30 M.
(a) What is the concentration of H_{2}O_{2} after 4.00 h?
(b) How long will it take for the H_{2}O_{2} concentration to drop to 0.12 M?
(c) How long will it take for 90% of the H_{2}O_{2} to decompose?
STRATEGY
Since this reaction has a first-order rate law, – \Delta [H_{2}O_{2}] / \Delta t = k [H_{2}O_{2}] , we can use the corresponding concentration–time equation for a first-order reaction:
\ln \frac{[H_{2}O_{2}]_{t}}{[H_{2}O_{2}]_{0}} = – ktIn 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 h) (\frac{60 min}{h}) (\frac{60 s}{min}) = 14,400 sThen, substitute the values of [H_{2}O_{2}]_{0}, k , and t into the concentration–time equation:
\ln \frac{[H_{2}O_{2}]_{t}}{0.30 M} = – (1.8 \times 10^{-5} s^{-1}) (1.44 \times 10^{4} s) = -0.259Taking the antilogarithm (antiln) of both sides gives
\frac{[H_{2}O_{2}]_{t}}{0.30 M} = e^{-0.259} = 0.772 \\ [H_{2}O_{2}]_{t} = (0.772) (0.30 M) = 0.23 M(b) First, solve the concentration–time equation for the time:
t = -\frac{1}{k} \ln \frac{[H_{2}O_{2}]_{t}}{[H_{2}O_{2}]_{0}}Then evaluate the time by substituting the concentrations and the value of k:
t = – (\frac{1}{1.8 \times 10^{-5} s^{-1}}) \ln \frac{0.12 M}{0.30 M} = -(\frac{1}{1.8 \times 10^{-5} s^{-1}}) (- 0.916) = 5.1 \times 10^{4} sThus, the H_{2}O_{2} concentration reaches 0.12 M at a time of 5.1 \times 10^{4} s (14 h).
(c) When 90% of the H_{2}O_{2} has decomposed, 10% remains. Therefore,
\frac{[H_{2}O_{2}]_{t}}{[H_{2}O_{2}]_{0}} = \frac{(0.10) (0.30 M)}{0.30 M} = 0.10The time required for 90% decomposition is
t = – (\frac{1}{1.8 \times 10^{-5} s^{-1}}) \ln 0.10 = – (\frac{1}{1.8 \times 10^{-5} s^{-1}}) (-2.30) = 1.3 \times 10^{5} s (36 h)✓BALLPARK CHECK The concentration of H_{2}O_{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 a still more time (36 h) is needed for the concen tration to fall to 0.030 M (10% of the original concentration). These results look rea sonable. A plot of H_{2}O_{2} versus time would exhibit an exponential decay of the H_{2}O_{2} concentration, as expected for a first-order reaction.