Read the passage carefully and answer the Questions.

The decomposition of $N_2O_5 (g)$ is given as

$N_2Os (g) → 2NO_2(g) +\frac{1}{2}O_2(g)$

The above reaction is found to be of first order.

Choose the correct statement about the rate of the reaction

The rate of the reaction doubles as the initial amount of $N_2O_5$ is doubled

The correct answer is Option (1) → The rate of the reaction doubles as the initial amount of $N_2O_5$ is doubled

The relationship between the rate and the concentration of reactants is defined by the Rate Law. For a first-order reaction, the rate is directly proportional to the concentration of the reactant raised to the power of one.

1. The Rate Law Expression:

$\text{Rate} = k[N_2O_5]^1$

Where:

• $k$ is the rate constant.
• $[N_2O_5]$ is the molar concentration of the reactant.

2. The Effect of Doubling Concentration:

If we change the initial concentration from $[N_2O_5]$ to $2 \times [N_2O_5]$, the new rate calculation is:

$\text{New Rate} = k(2 \times [N_2O_5]) = 2 \times (k[N_2O_5]) = 2 \times \text{Original Rate}$

Because the exponent (order) is 1, any change in the concentration results in an identical proportional change in the rate.

Why the other options are incorrect

• "Changing the amount has no effect": This describes a zero-order reaction, where $\text{Rate} = k[A]^0$.
• "Rate becomes half": This would imply an inverse relationship, which does not occur in standard simple reaction orders.
• "Order of reaction doubles with temperature": This is a common misconception. Temperature increases the rate constant ($k$) (often doubling the rate for every 10 °C rise, according to the Arrhenius equation), but the order of the reaction is determined by the mechanism and remains constant regardless of temperature.