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Chapter 14: Problem 24

The thermal decomposition of phosphine \(\left(\mathrm{PH}_{3}\right)\) into phosphorus and molecular hydrogen is a first-order reaction: $$ 4 \mathrm{PH}_{3}(g) \longrightarrow \mathrm{P}_{4}(g)+6 \mathrm{H}_{2}(g) $$ The half-life of the reaction is 35.0 s at \(680^{\circ} \mathrm{C}\). Calculate (a) the first-order rates constant for the reaction and (b) the time required for 95 percent of the phosphine to decompose.

Short Answer

Expert verified
The first-order rate constant for the reaction is \(0.0198\, \mathrm{s}^{-1}\) and the time required for 95 percent of the phosphine to decompose is 160 s.

Step by step solution

01

Calculation of First-Order Rate Constant

We have been given the half-life of the reaction, which is 35.0 s. We can now calculate the rate constant using the equation: \(k = \frac{0.693}{t_{1/2}}\). Plugging the given half-life into this equation gives \(k = \frac{0.693}{35.0\, \mathrm{s}} = 0.0198\, \mathrm{s}^{-1}\)
02

Calculation of Time for 95 Percent Decomposition

Next, we need to find out how much time it would take for 95% of the phosphine to decompose. This means that only 5% of the original amount remains. We use the integrated rate law for a first-order reaction: \(\ln\left(\frac{[A]_0}{[A]}\right) = kt\). Here, \([A]_0\) is the initial concentration and \([A]\) is the final concentration, which is 5% of \([A]_0\). Substituting \([A]_0\) and \([A]\) into the equation gives \(\ln\left( \frac{[A]_0}{0.05[A]_0}\right)\) which simplifies to \(\ln(20) = kt\). To find the time, we rearrange this equation and substitute the rate constant found in step 1: \(t = \frac{\ln(20)}{k}\) and finally \(t = \frac{\ln(20)}{0.0198\, \mathrm{s}^{-1}} = 160\, \mathrm{s}\).

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Most popular questions from this chapter

Consider the reaction $$ X+Y \longrightarrow Z $$ These data are obtained at \(360 \mathrm{~K}\): $$ \begin{aligned} &\text { Initial Rate of }\\\ &\begin{array}{ccc} \text { Disappearance of } \mathrm{X}(\mathrm{M} / \mathrm{s}) & {[\mathrm{X}]} & {[\mathrm{Y}]} \\ \hline 0.147 & 0.10 & 0.50 \\ 0.127 & 0.20 & 0.30 \\ 4.064 & 0.40 & 0.60 \\ 1.016 & 0.20 & 0.60 \\ 0.508 & 0.40 & 0.30 \end{array} \end{aligned} $$ (a) Determine the order of the reaction. (b) Determine the initial rate of disappearance of \(X\) when the concentration of \(\mathrm{X}\) is \(0.30 \mathrm{M}\) and that of \(\mathrm{Y}\) is \(0.40 \mathrm{M}\)

A certain first-order reaction is 35.5 percent complete in 4.90 min at \(25^{\circ} \mathrm{C}\). What is its rate constant?

The rate constant for the second-order reaction $$ 2 \mathrm{NOBr}(g) \longrightarrow 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) $$ is \(0.80 / M \cdot \mathrm{s}\) at \(10^{\circ} \mathrm{C}\). (a) Starting with a concentration of \(0.086 M,\) calculate the concentration of \(\mathrm{NOBr}\) after \(22 \mathrm{~s}\) (b) Calculate the half-lives when \([\mathrm{NOBr}]_{0}=\) \(0.072 M\) and \([\mathrm{NOBr}]_{0}=0.054 \mathrm{M}\).

Consider the zero-order reaction \(\mathrm{A} \longrightarrow\) product. (a) Write the rate law for the reaction. (b) What are the units for the rate constant? (c) Plot the rate of the reaction versus [A].

Consider the reaction $$ \mathrm{A}+\mathrm{B} \longrightarrow \text { products } $$ From these data obtained at a certain temperature, determine the order of the reaction and calculate the rate constant: $$ \begin{array}{ccc} {[\mathrm{A}](\mathrm{M})} & {[\mathrm{B}](M)} & \text { Rate }(M / \mathrm{s}) \\ \hline 1.50 & 1.50 & 3.20 \times 10^{-1} \\ 1.50 & 2.50 & 3.20 \times 10^{-1} \\ 3.00 & 1.50 & 6.40 \times 10^{-1} \end{array} $$
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