Question

The rate constant for the second-order reaction
2NO2(g) ---> 2NO(g) + O2(g)
is 0.345/M◦s at 275°C. How long (in s) would it take
for the concentration of NO2 to decrease from 0.60
M to 0.27 M?

Answer 1

We use the formula;

$\text{average rate}=\frac{\Delta \text{C}}{\Delta \text{t}}$

Rate constant = 0.345M^-1s^-1

$\Delta \text{C}$ = 0.60 - 0.27 = 0.33M

$\Delta \text{C}=\frac{0.33M}{0.345/Mxs}$

= 0.957s