Consider the following data for the given reaction
\(2\)\(\text{HI}_{(g)}\) \(\rightarrow\) \(\text{H}_2{(g)}\)$ + $\(\text{I}_2{(g)}\)
The order of the reaction is __________.
Consider the following data for the given reaction
\(2\)\(\text{HI}_{(g)}\) \(\rightarrow\) \(\text{H}_2{(g)}\)$ + $\(\text{I}_2{(g)}\)
The order of the reaction is __________.
Correct Answer: 2
Approach Solution - 1
To determine the order of the reaction, analyze the rate equation for a general reaction: \(te = k [\text{HI}]^n\).
Use the provided data to find the reaction order \(n\).
Compare experiments 1 and 2:
\((\frac{[\text{HI}]_2}{[\text{HI}]_1})^n = \frac{\text{Rate}_2}{\text{Rate}_1}\)
\(\left(\frac{0.01}{0.005}\right)^n = \frac{3.0 \times 10^{-3}}{7.5 \times 10^{-4}}\)
\(2^n = 4\)
Solve for \(n\): \(n = 2\).
Verify with experiments 2 and 3:
\(\left(\frac{0.02}{0.01}\right)^n = \frac{1.2 \times 10^{-2}}{3.0 \times 10^{-3}}\)
\(2^n = 4\)
This confirms \(n = 2\).
The reaction order is 2.
Approach Solution -2
Assuming the rate law:
$$\text{Rate} = k[\text{HI}]^n$$
Using any two of the given data points:
$$\frac{3.0 \times 10^{-3}}{7.5 \times 10^{-4}} = \left(\frac{0.01}{0.005}\right)^n$$
Solving, we find \( n = 2 \), so the reaction is second order.
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