Question

For the reaction $ A \rightarrow $ products, 

The reaction was started with 2.5 mol L\(^{-1}\) of A.

Hint:

In a zero-order reaction, the rate of reaction is constant and the concentration of reactant decreases linearly with time. The equation \( [A] = -Kt + A_0 \) is used to calculate the concentration at any given time.

Answer 1

Correct Answer: 2435

The graph indicates that \( t_{1/2} \) is directly proportional to \( [A] \). The provided slope is 76.92. For a zero-order reaction, the relationship is \( t_{1/2} = \frac{A_0}{2K} \), where the slope is \( \frac{1}{2K} = 76.92 \). Consequently, \( K = \frac{1}{2 \times 76.92} = \frac{1}{153.84} \). Using the zero-order reaction equation, \( [A] = - Kt + A_0 \), the concentration of A at 10 minutes is calculated as \( [A] = - \frac{1}{2 \times 76.92} \times 10 + 2.5 = 2.435 \, \text{mol/L} \). Therefore, the concentration of A at 10 minutes is \( 2435 \times 10^{-3} \, \text{mol/L} \).