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
Answer & explanation
Correct answer: option 1
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.