JEE Chemistry 2025 Question with Solution

Given: Initial total pressure at $t = 0$ is $0.6 \, \text{atm}$, rate constant $k = 4.606 \times 10^{-2} \, s^{-1}$, and time $t = 100 \, \text{s}$.

Find: The value of $x$ in total pressure at $100 \, \text{s}$ for the reaction

$$2N_2O_5(g) \rightarrow 2N_2O_4(g) + O_2(g)$$

For a first-order reaction,

$$k = \frac{1}{t} \ln\left(\frac{P_0}{P_t}\right)$$

where $P_0$ is the initial pressure of $N_2O_5$ and $P_t$ is its pressure at time $t$.

Substituting the given values,

$$4.606 \times 10^{-2} = \frac{1}{100} \ln\left(\frac{0.6}{P_t}\right)$$ $$\ln\left(\frac{0.6}{P_t}\right) = 4.606$$ $$\frac{0.6}{P_t} = e^{4.606} \approx 100$$ $$P_t = \frac{0.6}{100} = 0.006 \, \text{atm}$$

Let the pressure of $O_2$ formed be $p$. Then from stoichiometry, the decrease in pressure of $N_2O_5$ is $2p$.

So,

$$0.6 - 2p = 0.006$$ $$2p = 0.594$$ $$p = 0.297$$

At time $t$, total pressure is

$$P_{\text{total}} = (0.6 - 2p) + 2p + p = 0.6 + p$$ $$P_{\text{total}} = 0.6 + 0.297 = 0.897 \, \text{atm}$$

Therefore,

$$x = 897 \times 10^{-3} \, \text{atm}$$

So the required nearest integer is 897.

The solution shows a discrepancy with the answer key, but the worked solution gives 897.

Given: Initially only $N_2O_5$ is present with total pressure $0.6 \, \text{atm}$.

Find: Total pressure after $100 \, \text{s}$.

Let pressure of $O_2$ formed after $100 \, \text{s}$ be $p$. From

$$2N_2O_5(g) \rightarrow 2N_2O_4(g) + O_2(g)$$

if $O_2$ formed is $p$, then:

• pressure decrease of $N_2O_5$ is $2p$
• pressure increase of $N_2O_4$ is $2p$

Hence pressure of $N_2O_5$ left is $0.6 - 2p$.

Using first-order kinetics,

$$k = \frac{1}{t} \ln\left(\frac{[N_2O_5]_0}{[N_2O_5]_t}\right) = \frac{1}{t} \ln\left(\frac{0.6}{0.6 - 2p}\right)$$

Substitute the values:

$$4.606 \times 10^{-2} = \frac{1}{100} \ln\left(\frac{0.6}{0.6 - 2p}\right)$$ $$\ln\left(\frac{0.6}{0.6 - 2p}\right) = 4.606$$ $$\frac{0.6}{0.6 - 2p} = e^{4.606} \approx 100$$ $$0.6 - 2p = \frac{0.6}{100} = 0.006$$ $$2p = 0.594$$ $$p = 0.297$$

Now total pressure at time $t$ is

$$P(t) = (0.6 - 2p) + 2p + p = 0.6 + p$$ $$P(t) = 0.6 + 0.297 = 0.897 \, \text{atm}$$

Therefore, $x = 897 \times 10^{-3} \, \text{atm}$ and the numerical answer is 897.