$$r = kA$$ for a reaction, 50% of A is decomposed in 120 minutes. The time taken for 90% decomposition of A is ______ minutes.

The rate expression $$r = kA$$ shows that the rate is directly proportional to the concentration of $$A$$, so the reaction follows first-order kinetics.

For a first-order reaction the integrated rate law is
$$\ln\!\left(\frac{[A]_0}{[A]}\right)=kt$$ $$-(1)$$
and the half-life is related to the rate constant by
$$t_{1/2} = \frac{0.693}{k}$$ $$-(2)$$

Case 1: 50 % decomposition

50 % decomposition means $$[A] = \tfrac{1}{2}[A]_0$$. By definition this is the half-life, so
$$t_{1/2}=120\ \text{min}$$

Using $$(2)$$,
$$k = \frac{0.693}{t_{1/2}} = \frac{0.693}{120\ \text{min}} = 0.005775\ \text{min}^{-1}$$

Case 2: 90 % decomposition

90 % decomposition leaves $$10\%$$ of the initial concentration, so $$[A]=0.1[A]_0$$.
Substitute into $$(1)$$:

$$\ln\!\left(\frac{[A]_0}{0.1[A]_0}\right)=kt$$
$$\ln 10 = k\,t$$
$$2.303 = k\,t$$

Solve for $$t$$:
$$t = \frac{2.303}{k}= \frac{2.303}{0.005775\ \text{min}^{-1}} = 3.992\times10^{2}\ \text{min} \approx 399\ \text{min}$$

Therefore, the time required for 90 % decomposition of $$A$$ is about 399 minutes.