Question

Write unit of first order reaction. The initial concentration of N$_2$O$_5$ at 318 K was $0.60\times 10^{-2$ mol L$^{-1}$ in the first order reaction, which became $0.20\times 10^{-2}$ mol L$^{-1}$ after 60 min. Calculate the velocity constant at 318 K. (log 3 = 0.4771).}

Hint:

For first order, plot or use \(k=\dfrac{2.303}{t}\log\!\left(\dfrac{a}{a_t}\right)\).
When concentration drops by a factor of 3 in time \(t\), simply use \(\log 3\).

Answer 1

Unit of a first-order rate constant:
Time$^{-1}$ (e.g., s$^{-1}$ or min$^{-1}$).
Data:
\(a = 0.60\times 10^{-2} = 6.0\times 10^{-3}\ \text{mol L}^{-1}\).
\(a_t = 0.20\times 10^{-2} = 2.0\times 10^{-3}\ \text{mol L}^{-1}\).
\(t = 60\ \text{min}\).
First-order relation:
\(k = \dfrac{2.303}{t}\log\!\left(\dfrac{a}{a_t}\right) = \dfrac{2.303}{60}\log(3).\)
Given \(\log 3 = 0.4771\):
\(k = \dfrac{2.303\times 0.4771}{60} = \dfrac{1.099}{60} = 1.831\times 10^{-2}\ \text{min}^{-1}.\)
In s$^{-1}$: \(k = \dfrac{1.831\times 10^{-2}}{60} = 3.05\times 10^{-4}\ \text{s}^{-1}.\)
\[ \boxed{k \approx 1.83\times 10^{-2}\ \text{min}^{-1}\; (\,=\,3.05\times 10^{-4}\ \text{s}^{-1}\,)} \]