Question

a chemistry graduate student is studying the rate of this reaction: 2hi(g)→h₂(g)+i₂(g). she fills a reaction vessel with hi and measures its concentration as the reaction proceeds: time (seconds) hi 0 0.600m 0.10 0.170m 0.20 0.0990m 0.30 0.0698m 0.40 0.0540m. use this data to answer the following questions. write the rate law for this reaction. rate = k . calculate the value of the rate constant k. round your answer to 2 significant digits. also be sure your answer has the correct unit symbol.

Explanation:

Step1: Determine reaction order

To find the reaction order with respect to HI, we can check the integrated - rate laws for different orders. For a second - order reaction, the integrated rate law is $\frac{1}{[A]_t}-\frac{1}{[A]_0}=kt$. Let's check if the reaction is second - order.
For $t = 0.10\ s$, $[HI]_0=0.600\ M$, $[HI]_t = 0.170\ M$.
$\frac{1}{[HI]_t}-\frac{1}{[HI]_0}=\frac{1}{0.170}-\frac{1}{0.600}=\frac{0.600 - 0.170}{0.170\times0.600}=\frac{0.430}{0.102}\approx4.22$.
For $t = 0.20\ s$, $[HI]_0 = 0.600\ M$, $[HI]_t=0.0990\ M$.
$\frac{1}{[HI]_t}-\frac{1}{[HI]_0}=\frac{1}{0.0990}-\frac{1}{0.600}=\frac{0.600 - 0.0990}{0.0990\times0.600}=\frac{0.501}{0.0594}\approx8.43$.
The slopes of $\frac{1}{[HI]_t}$ vs $t$ are approximately constant, so the reaction is second - order with respect to HI. The rate law is rate = $k[HI]^2$.

Step2: Calculate the rate constant $k$

Using the integrated rate - law $\frac{1}{[HI]_t}-\frac{1}{[HI]_0}=kt$. We can take any two data points. Let's take the first two points: $t = 0.10\ s$, $[HI]_0 = 0.600\ M$, $[HI]_t=0.170\ M$.
$k=\frac{\frac{1}{[HI]_t}-\frac{1}{[HI]_0}}{t}$
$k=\frac{\frac{1}{0.170}-\frac{1}{0.600}}{0.10}=\frac{\frac{0.600 - 0.170}{0.170\times0.600}}{0.10}=\frac{\frac{0.430}{0.102}}{0.10}=\frac{4.22}{0.10}=42\ M^{-1}s^{-1}$

Answer:

rate = $k[HI]^2$
$k = 42\ M^{-1}s^{-1}$