A (initial concentration $0.800 M$ ) undergoes parallel first-order reactions $A B$ ( $k_{1} = 0.030 mi n^{- 1}$ ) and $A C$ ( $k_{2} = 0.010 mi n^{- 1}$ ). After $t = 34.65$ min, find $[C]$ in M (two decimals). Use $e^{- 1.386} = 0.250$ .

Numerical answer type, answer in M, tolerance 0.01.

Answer: 0.15 M

Here

(k_{1} + k_{2}) t = 0.040 \times 34.65 = 1.386

, so

[A] = [A]_{0} e^{- 1.386} = 0.250 \times 0.800 = 0.200 M

and

0.600 M

has reacted. C receives the fraction

k_{2} / (k_{1} + k_{2}) = 0.25

of the reacted A:

[C] = 0.25 \times 0.600 = 0.15 M