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Given the reaction a → b, if the rate constant k is 0.25 mol L^{-1} s^{-1} at 75 °C and the reaction is first order in a, what is the correct rate law for this reaction?
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15. Chemical Kinetics
Rate Law
Scelta multipla
Using the information in the table below, determine the rate law for the reaction: a(g) + 3 b(g) → c(g) + 2 d(g). | Experiment | [a] (M) | [b] (M) | Initial Rate (M/s) | |------------|--------|--------|--------------------| | 1 | 0.10 | 0.10 | 0.020 | | 2 | 0.20 | 0.10 | 0.040 | | 3 | 0.10 | 0.20 | 0.080 | Which of the following is the correct rate law?
A
rate = k [a]^1 [b]^1
B
rate = k [a]^0 [b]^2
C
rate = k [a]^1 [b]^2
D
rate = k [a]^2 [b]^1
Guida verificata passo dopo passo1
Write the general form of the rate law for the reaction: \(\text{rate} = k [a]^m [b]^n\), where \(m\) and \(n\) are the orders of the reaction with respect to reactants \(a\) and \(b\), respectively.
Compare experiments 1 and 2 to determine the order with respect to \(a\). Since \([b]\) is constant at 0.10 M, observe how the rate changes when \([a]\) doubles from 0.10 M to 0.20 M. Use the ratio of rates to find \(m\) by setting up the equation: \(\frac{\text{rate}_2}{\text{rate}_1} = \left(\frac{[a]_2}{[a]_1}\right)^m\).
Compare experiments 1 and 3 to determine the order with respect to \(b\). Since \([a]\) is constant at 0.10 M, observe how the rate changes when \([b]\) doubles from 0.10 M to 0.20 M. Use the ratio of rates to find \(n\) by setting up the equation: \(\frac{\text{rate}_3}{\text{rate}_1} = \left(\frac{[b]_3}{[b]_1}\right)^n\).
Solve the equations from steps 2 and 3 to find the values of \(m\) and \(n\), which represent the reaction orders with respect to \(a\) and \(b\), respectively.
Write the final rate law by substituting the determined reaction orders \(m\) and \(n\) into the general rate law expression: \(\text{rate} = k [a]^m [b]^n\).
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