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For the reaction 2NO + O2 → 2NO2, the experimentally determined rate law is rate = k[NO]^2[O2]. What is the overall reaction order, what is the order with respect to O2, and what are the units of the rate constant k?
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15. Chemical Kinetics
Rate Law
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Given the reaction 2NO + O_2 → 2NO_2 and the following initial rate data: | [NO] (M) | [O_2] (M) | Initial Rate (M/s) | |----------|-----------|---------------------| | 0.10 | 0.10 | 1.2 × 10^{-3} | | 0.20 | 0.10 | 4.8 × 10^{-3} | | 0.10 | 0.20 | 2.4 × 10^{-3} | Determine the correct rate law and value of the rate constant k.
A
Rate = k[NO][O_2]^2, k = 6.0 × 10^1 M^{-2}s^{-1}
B
Rate = k[NO][O_2], k = 1.2 × 10^{-2} M^{-1}s^{-1}
C
Rate = k[NO]^2, k = 1.2 × 10^{-3} M^{-2}s^{-1}
D
Rate = k[NO]^2[O_2], k = 1.2 × 10^2 M^{-2}s^{-1}
Guida verificata passo dopo passo1
Write the general form of the rate law for the reaction: \(\text{Rate} = k[\text{NO}]^m[\text{O}_2]^n\), where \(m\) and \(n\) are the reaction orders with respect to NO and O\(_2\), respectively.
Use the initial rate data to determine the order with respect to NO by comparing experiments where [O\(_2\)] is constant. For example, compare experiment 1 and 2 where [O\(_2\)] = 0.10 M:
\[\frac{\text{Rate}_2}{\text{Rate}_1} = \frac{k[0.20]^m[0.10]^n}{k[0.10]^m[0.10]^n} = \left(\frac{0.20}{0.10}\right)^m = 4.8 \times 10^{-3} / 1.2 \times 10^{-3} = 4.\]
Solve for \(m\) by taking the logarithm: \(4 = 2^m\) implies \(m = 2\).
Next, determine the order with respect to O\(_2\) by comparing experiments where [NO] is constant, such as experiment 1 and 3 where [NO] = 0.10 M:
\[\frac{\text{Rate}_3}{\text{Rate}_1} = \frac{k[0.10]^m[0.20]^n}{k[0.10]^m[0.10]^n} = \left(\frac{0.20}{0.10}\right)^n = 2.4 \times 10^{-3} / 1.2 \times 10^{-3} = 2.\]
Solve for \(n\) by taking the logarithm: \(2 = 2^n\) implies \(n = 1\).
Write the rate law with the determined orders: \(\text{Rate} = k[\text{NO}]^2[\text{O}_2]^1\).
Use any set of data to solve for the rate constant \(k\). For example, using experiment 1:
\[k = \frac{\text{Rate}}{[\text{NO}]^2[\text{O}_2]} = \frac{1.2 \times 10^{-3}}{(0.10)^2 \times 0.10}.\]
Calculate \(k\) (do not provide the final numeric value here) and include the units based on the rate law orders: \(\text{units of } k = M^{-(m+n-1)} s^{-1} = M^{-2} s^{-1}\).
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