A substance dissociates through a first-order reaction with a half-life of 32 ext min. If the initial concentration of the substance is 2.7 imes 1010 ext molecules/L, what will be the concentration 2.

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

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

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:

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

Where $[A]_t$ is the concentration at time $t$, and $[A]_0$ is the initial concentration. This can also be written as:

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

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

$$[A]_t = \boxed{1.5 \times 10^9 \text{ molecules/L}}$$

ii) Express the rate of the following reaction in terms of the formation of ammonia: $\text{N}_2(\text{g}) + 3\text{H}_2(\text{g}) \to 2\text{NH}_3(\text{g})$.

The rate of reaction is expressed as the change in concentration of a reactant or product over time, divided by its stoichiometric coefficient. Reactants have a negative sign because their concentration decreases.

$$\text{Rate} = -\frac{d[\text{N}_2]}{dt} = -\frac{1}{3}\frac{d[\text{H}_2]}{dt} = \frac{1}{2}\frac{d[\text{NH}_3]}{dt}$$

The formation of ammonia is represented by the term $\frac{d[\text{NH}_3]}{dt}$. To express the overall rate of reaction in terms of this, we use the coefficient.

$$\text{Rate} = \boxed{\frac{1}{2}\frac{d[\text{NH}_3]}{dt}}$$