N$$_2$$O$$_{5(g)} \to$$ 2NO$$_{2(g)}$$ + $$\frac{1}{2}$$O$$_{2(g)}$$
In the above first order reaction the initial concentration of N$$_2$$O$$_5$$ is $$2.40 \times 10^{-2}$$ mol L$$^{-1}$$ at 318 K. The concentration of N$$_2$$O$$_5$$ after 1 hour was $$1.60 \times 10^{-2}$$ mol L$$^{-1}$$. The rate constant of the reaction at 318 K is ___ $$\times 10^{-3}$$ min$$^{-1}$$ (Nearest integer):
[Given : log 3 = 0.477, log 5 = 0.699]

For a first-order reaction, the rate constant is given by $$k = \frac{2.303}{t}\log\frac{[\text{A}]_0}{[\text{A}]}$$.

Here $$[\text{A}]_0 = 2.40 \times 10^{-2}$$ mol/L, $$[\text{A}] = 1.60 \times 10^{-2}$$ mol/L, and $$t = 1 \text{ hour} = 60 \text{ min}$$.

The ratio is $$\frac{[\text{A}]_0}{[\text{A}]} = \frac{2.40}{1.60} = \frac{3}{2}$$. So $$\log\frac{3}{2} = \log 3 - \log 2 = 0.477 - 0.301 = 0.176$$. Note that $$\log 2 = \log\frac{10}{5} = \log 10 - \log 5 = 1 - 0.699 = 0.301$$.

Therefore $$k = \frac{2.303}{60} \times 0.176 = \frac{0.4053}{60} = 6.76 \times 10^{-3} \text{ min}^{-1}$$.

Expressed as $$\_\_ \times 10^{-3} \text{ min}^{-1}$$, the nearest integer is $$7$$.