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33 lines
1.2 KiB
Markdown
33 lines
1.2 KiB
Markdown
# Problem: 03-02-2
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## Question
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The rate law for a particular reaction is rate $=k[\mathrm{XY}]^2$. In an experiment, the initial rate of the reaction is determined to be $0.16 \mathrm{~mol} /(\mathrm{L} \cdot \mathrm{s})$ when the initial concentration of $\mathrm{XY}$ is $0.40 \mathrm{~mol} / \mathrm{L}$.
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- What is the value of the rate constant, $k$, for the reaction?
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- [ ] $0.10 \mathrm{~L} /(\mathrm{mol} \cdot \mathrm{s})$
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- [ ] $0.40 \mathrm{~L} /(\mathrm{mol} \cdot \mathrm{s})$
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- [x] $1.0 \mathrm{~L} /(\mathrm{mol} \cdot \mathrm{s})$
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- [ ] $2.5 \mathrm{~L} /(\mathrm{mol} \cdot \mathrm{s})$
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## Solution
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lo tind the value of the rate constant for the reaction, let's first solve the rate law for $k$ :
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$$
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k=\frac{\text { rate }}{[\mathrm{XY}]^2}
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$$
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Next, let's plug in the initial rate and concentration given in the text:
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$$
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\begin{aligned}
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k & =\frac{0.16 \mathrm{~mol} /(\mathrm{L} \cdot \mathrm{s})}{(0.40 \mathrm{~mol} / \mathrm{L})^2} \\
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& =\frac{0.16 \mathrm{~mol} /(\mathrm{L} \cdot \mathrm{s})}{0.16 \mathrm{~mol}^2 / \mathrm{L}^2} \\
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& =1.0 \mathrm{~L} /(\mathrm{mol} \cdot \mathrm{s})
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\end{aligned}
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$$
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So, the value of the rate constant for the reaction is
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$1.0 \mathrm{~L} /(\mathrm{mol} \cdot \mathrm{s})$
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