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Chemical Kinetics and Nuclear Chemistry question

2025 · 24 Jan · Shift 1 · Q1

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Chemical Kinetics and Nuclear Chemistry
2025 · 24 Jan · Shift 1 · Q1

Chemical Kinetics and Nuclear Chemistry question

2025 · 24 Jan · Shift 1 · Q1

JEE MainChemistryChemical Kinetics and Nuclear ChemistryMCQ+4 / −1

For a reaction,

\mathrm{N}_2 \mathrm{O}_{5(\mathrm{~g})} \rightarrow 2 \mathrm{NO}_{2(\mathrm{~g})}+\frac{1}{2} \mathrm{O}_{2(\mathrm{~g})}

in a constant volume container, no products were present initially. The final pressure of the system when

50 \%

of reaction gets completed is

A
$5 / 2$ times of initial pressure
B
$7 / 2$ times of initial pressure
C
$7 / 4$ times of initial pressure
D
5 times of initial pressure

View written solutionFree

Correct answer: C

Given reaction

$\mathrm{N_2O_5(g)} \rightarrow 2\,\mathrm{NO_2(g)} + \frac{1}{2}\,\mathrm{O_2(g)}$

The container has constant volume and we assume temperature is unchanged, so for gases:

$P \propto n$

That means pressure is directly proportional to total moles of gas.

Assume initial moles

Let initial moles of $\mathrm{N_2O_5}$ be $1$ mol.

Initially:

$\mathrm{N_2O_5} = 1$
$\mathrm{NO_2} = 0$
$\mathrm{O_2} = 0$

So initial total moles:

$n_i = 1$

Let initial pressure be $P_i$ .

At 50% completion of reaction

$50\%$ of $\mathrm{N_2O_5}$ reacts, so reacted moles = $0.5$ mol.

From stoichiometry:

$1\,\mathrm{N_2O_5} \rightarrow 2\,\mathrm{NO_2} + \frac{1}{2}\,\mathrm{O_2}$

Therefore, if $0.5$ mol reacts:

$\mathrm{N_2O_5}$ left: $1 - 0.5 = 0.5$
$\mathrm{NO_2}$ formed: $2 \times 0.5 = 1.0$
$\mathrm{O_2}$ formed: $\frac{1}{2} \times 0.5 = 0.25$

So total moles at this stage:

$n_f = 0.5 + 1.0 + 0.25 = 1.75 = \frac{7}{4}$

Relate pressure to moles

Since volume is constant,

$\frac{P_f}{P_i} = \frac{n_f}{n_i} = \frac{7/4}{1} = \frac{7}{4}$

Thus,

$P_f = \frac{7}{4} P_i$

Match with options

The final pressure is:

$\boxed{\frac{7}{4} \text{ times the initial pressure}}$

So the correct option is C.

Comparison with stored answer

Stored correct answer: C

Our derived answer: C

Hence, they agree.

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