For the given reaction, if the initial pressure is 450 mmHg and the pressure at time t is 720 mmHg at a constant temperature T and constant volume V. The fraction of A( g) decomposed under these…

Given: The reaction is A( g) 2B( g) + C( g) Initial pressure of A is 450 mmHg. Total pressure at time t is 720 mmHg at constant temperature and constant volume. Find: The value of x in the fraction decomposed written as x 10^-1. Let the pressure decrease of A be y. Then at time t, PA = 450 - y, PB = 2y, PC = y So total pressure is Pt = (450 - y) + 2y + y Using the given total pressure, 720 = 450 + 2y 270 = 2y y = 135 Therefore, the fraction of A decomposed is y450 = 135450 = 0.3 = 3 10^-1 Hence, the value of x is 3.

For the given reaction, if the initial pressure is 450 mmHg and t… — Solution | JEE Chemistry

NVAEasyJEE 2023Stoichiometry & Calculations

Reaction scheme showing gaseous species A in equilibrium with 2B plus C, all in gaseous state.

For the given reaction, if the initial pressure is $450 \, \text{mmHg}$ and the pressure at time $t$ is $720 \, \text{mmHg}$ at a constant temperature $T$ and constant volume $V$ . The fraction of $A(\text{g})$ decomposed under these conditions is $x \times 10^{-1}$ . The value of $x$ is _____ (nearest integer)

Answer

Correct answer:3

Step-by-step solution

Standard Method

Given: The reaction is

A(\text{g}) \rightleftharpoons 2B(\text{g}) + C(\text{g})

Initial pressure of $A$ is $450 \, \text{mmHg}$ . Total pressure at time $t$ is $720 \, \text{mmHg}$ at constant temperature and constant volume.

Find: The value of $x$ in the fraction decomposed written as $x \times 10^{-1}$ .

Let the pressure decrease of $A$ be $y$ . Then at time $t$ ,

P_A = 450 - y, \quad P_B = 2y, \quad P_C = y

So total pressure is

P_t = (450 - y) + 2y + y

Using the given total pressure,

720 = 450 + 2y

270 = 2y

y = 135

Therefore, the fraction of $A$ decomposed is

\frac{y}{450} = \frac{135}{450} = 0.3 = 3 \times 10^{-1}

Hence, the value of $x$ is $3$ .

Common mistakes

Taking the decomposed amount of $A$ as the same as the total pressure increase is incorrect because products contribute more moles to the total pressure. First write partial pressures using stoichiometry, then add them to match the total pressure.
Using the final pressure $720 \, \text{mmHg}$ as the pressure of $A$ alone is wrong because it is the total pressure of all gases present. Always separate $P_A$ , $P_B$ , and $P_C$ before solving.
Computing the fraction decomposed as $\frac{135}{720}$ is incorrect because decomposition fraction is based on the initial amount of A, not the final total pressure. Use $\frac{\text{amount of A decomposed}}{\text{initial amount of A}} = \frac{135}{450}$ .

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