The decomposition of nitrous oxide at 565 ºC
N $_{2}$ O(g) → N $_{2}$ (g) + ½ O $_{2}$ (g)
is second order in N $_{2}$ O with a rate constant of 1.10 × 10 $^{-3}$ M $^{-1}$ s $^{-1}$ . If an experiment is performed where the initial concentration of N $_{2}$ O is 0.108 M, what is the N $_{2}$ O concentration after 1250 seconds?

Question

The decomposition of nitrous oxide at 565 ºC
N[latex]_{2}[/latex]O(g) → N[latex]_{2}[/latex](g) + ½ O[latex]_{2}[/latex](g)
is second order in N[latex]_{2}[/latex]O with a rate constant of 1.10 × 10[latex]^{-3}[/latex] M[latex]^{-1}[/latex]s[latex]^{-1}[/latex]. If an experiment is performed where the initial concentration of N[latex]_{2}[/latex]O is 0.108 M, what is the N[latex]_{2}[/latex]O concentration after 1250 seconds?

Accepted Answer

You are asked to calculate the concentration of a reactant after a given period of time.
You are given the balanced equation for the reaction, the order of the reaction with respect to the reactant, the rate constant for the reaction, the initial concentration of the reactant, and the period of time.
Because the reaction is second order, use the second-order integrated rate law.

[latex]\frac{1}{\left[N_{2}O ight]_{t} }= \frac{1}{\left[N_{2}O ight]_{0} }+kt[/latex] t = 1250 seconds

[latex]\left[N_{2}O ight]_{0}[/latex] = 0.108 M k = 1.10 × 10[latex]^{−3}M^{−1}s^{−1}[/latex]

[latex]\frac{1}{\left[N_{2}O ight]_{t} } = \frac{1}{0.108 ext{ M}}+(1.10 imes 10^{-3} M^{-1}s^{-1})[/latex](1250 s)

[latex]\frac{1}{\left[N_{2}O ight]_{t} }=10.6 ext{ M}^{-1}[/latex]

[latex]\left[N_{2}O ight]_{t}[/latex] = 0.0940 M

Chapter 14

Q. 14.4.1

Q. 14.4.1

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