For a first-order reaction, calculate the concentration of a substance after 2.24 hours, given an initial concentration of 2.7 x 1010 molecules/L and a half-life of 32 minutes.

The half-life is given in minutes, so we convert the total time from hours to minutes.

$$t = 2.24 \text{ h} \times 60 \frac{\text{min}}{\text{h}} = 134.4 \text{ min}$$

For a first-order reaction, the half-life ($t_{1/2}$) is related to the rate constant ($k$) by the formula:

$$t_{1/2} = \frac{\ln(2)}{k}$$

Rearranging to solve for $k$:

$$k = \frac{\ln(2)}{t_{1/2}}$$ $$k = \frac{0.693}{32 \text{ min}}$$ $$k \approx 0.02166 \text{ min}^{-1}$$

The integrated rate law is:

$$\ln([A]_t) = \ln([A]_0) - kt$$

Alternatively, in exponential form:

$$[A]_t = [A]_0 e^{-kt}$$

Where $[A]_t$ is the concentration at time $t$, and $[A]_0$ is the initial concentration.

$$[A]_t = (2.7 \times 10^{10} \text{ molecules/L}) \times e^{-(0.02166 \text{ min}^{-1} \times 134.4 \text{ min})}$$ $$[A]_t = (2.7 \times 10^{10} \text{ molecules/L}) \times e^{-2.9076}$$ $$[A]_t = (2.7 \times 10^{10} \text{ molecules/L}) \times 0.05462$$ $$[A]_t = 1.47474 \times 10^9 \text{ molecules/L}$$