Question

a certain reaction is first order in $ce{h_2}$ and first order in $ce{i_2}$. use this information to complete the table below. round each of your answer

$ce{h_2}$

$ce{i_2}$

initial rate of reaction

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1.76 $m$

0.946 $m$

0.688 $m/s$

2.38 $m$

0.946 $m$

$square$ $m/s$

0.571 $m$

2.92 $m$

$square$ $m/s$

Explanation:

Step1: Write rate law equation

Since the reaction is first order in $\text{H}_2$ and first order in $\text{I}_2$, the rate law is:
$$\text{Rate} = k[\text{H}_2][\text{I}_2]$$

Step2: Solve for rate constant $k$

Use the first row of data to find $k$:
$$k = \frac{\text{Rate}}{[\text{H}_2][\text{I}_2]} = \frac{0.688\ \text{M/s}}{(1.76\ \text{M})(0.946\ \text{M})}$$
$$k = \frac{0.688}{1.66496} \approx 0.4132\ \text{M}^{-1}\text{s}^{-1}$$

Step3: Calculate rate for row 2

Substitute $k$, $[\text{H}_2]=2.38\ \text{M}$, $[\text{I}_2]=0.946\ \text{M}$ into rate law:
$$\text{Rate} = (0.4132)(2.38)(0.946)$$
$$\text{Rate} \approx 0.4132 \times 2.25148 \approx 0.930\ \text{M/s}$$

Step4: Calculate rate for row 3

Substitute $k$, $[\text{H}_2]=0.571\ \text{M}$, $[\text{I}_2]=2.92\ \text{M}$ into rate law:
$$\text{Rate} = (0.4132)(0.571)(2.92)$$
$$\text{Rate} \approx 0.4132 \times 1.66732 \approx 0.689\ \text{M/s}$$

Answer:

Row 2 initial rate: $0.930\ \text{M/s}$
Row 3 initial rate: $0.689\ \text{M/s}$