TG EAMCET 3 May 2025 Shift 2 Paper

This problem involves a zero-order reaction, where the rate of reaction is independent of the reactant's concentration.The integrated rate law for a zero-order reaction is given by $[A]_t=[A]_0 - k t$ where,$[A]_t$ is the concentration at time $t$.$[A]_0$ is the initial concentration.$k$ is the rate constant.$t$ is time.The half-life ( $t_{1/2}$ ) for a zero-order reaction is related to the initial concentration and rate constant by the expression$t_{1/2} = \; \frac{[A]_0}{2k}$Given$t_{1/2} = 0.5 \text{h}$Initial concentration for half-life, $[A]_0 = 4 \mathrm{mol} \mathrm{L}^{-1}$.First, calculate the rate constant ( $k$ )$0.5 \text{hr} \; = \; \frac{4 \mathrm{mol} \mathrm{L}^{-1}}{2k}$$k \; = \; \frac{4 \mathrm{mol} \mathrm{L}^{-1}}{2 \times 0.5 \text{hr}} \; = \; \frac{4 \mathrm{mol} \mathrm{L}^{-1}}{1.0 \text{hr}}$$\; = 4 \mathrm{mol} \mathrm{L}^{-1} \text{hr}^{-1}$Next, calculate the time $(t)$ for the concentration to change from $2.0 \mathrm{mol} \mathrm{L}^{-1}$to $1.0 \mathrm{mol} \mathrm{L}^{-1}$;Using the integrated rate law$\; 1.0 \mathrm{mol} \mathrm{L}^{-1} = 2.0 \mathrm{mol} \mathrm{L}^{-1}$$\; - (4 \mathrm{mol} \mathrm{L}^{-1} \text{hr}^{-1}) t$ Rearrange to solve for $t$$\; (4 \mathrm{mol} \mathrm{L}^{-1} \text{hr}^{-1} t = 2.0 \mathrm{mol} \mathrm{L}^{-1} - 1.0 \mathrm{mol} \mathrm{L}^{-1})$$\; 4 t = 1.0 \text{hr}$$\; t = \; \frac{1.0}{4} \text{hr}$$\; t = \frac{1}{4} \text{h} \; \text{ or } \; 0.25 \text{h}.$