The gas-phase reaction N2O5 → 2NO2+1/2O2 is found to be first order. The rate constant at 337.6K is equal to 5.12×10^−3 s^−1.^a If the partial pressure of N2O5 is 0.500 atm at time t=0, find the partial pressure of N2O5 at t=60.0 s, neglecting any reverse reaction and assuming that the system is

Question

The gas-phase reaction

[latex] N_{2}O_{5} → 2NO_{2}+\frac{1}{2} O_{2} [/latex]

is found to be first order. The rate constant at 337.6K is equal to [latex]5.12×10^{−3} s^{−1}.^{a}[/latex] If the partial pressure of [latex]N_{2}O_{5}[/latex] is 0.500 atm at time t=0, find the partial pressure of [latex]N_{2}O_{5}[/latex] at t=60.0 s, neglecting any reverse reaction and assuming that the system is at constant volume.

Accepted Answer

Assuming the gas to be ideal, the partial pressure of a gas is proportional to the concentration. In the case of constant temperature [latex] Eq. (11.2-5): [A]_{t}= [A]_{0}e^{−k_{f}t} (first order, no reverse reaction) [/latex] can be rewritten in terms of pressure:

[latex]P(N_{2}O_{5})_{t}= P(N_{2}O_{5})_{0}e^{−k_{f}t}[/latex]

[latex]=(0.500 atm)exp[-(5.12×10^{−3} s^{−1})(60.0 s)]=0.368 atm [/latex]

[latex]^{a}[/latex] H. S. Johnston and Y. Tao, J. Am. Chem. Soc., 73, 2948 (1951).

Question 11.1: The gas-phase reaction N2O5 → 2NO2+1/2O2 is found to be firs...

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