Question

A gaseous compound decomposes on heating as per the following equation:

$A(g) \rightarrow B(g) + 2C(g)$. After 5 minutes and 20 seconds, the pressure increases by 96 mm Hg. If the rate constant for this first order reaction is 5.2 x $10^{-4} s^{-1}$, the initial pressure of A (in mm Hg) is: (Nearest Integer)

(Use: In(1.181) = 0.1664)

Answer 1

313

Answer 2

The decomposition reaction is given by:

$A(g) \rightarrow B(g) + 2C(g)$

Let $P_0$ be the initial pressure of A. At time t, let $x$ be the pressure of A that has decomposed. The pressures of the components at time t will be:

Pressure of A = $P_A = P_0 - x$

Pressure of B = $P_B = x$

Pressure of C = $P_C = 2x$

The total pressure at time t, $P_t$, is the sum of the partial pressures:

$P_t = P_A + P_B + P_C = (P_0 - x) + x + 2x = P_0 + 2x$

The increase in pressure is given as 96 mm Hg.

Increase in pressure = $P_t - P_0 = (P_0 + 2x) - P_0 = 2x$

Given, $2x = 96$ mm Hg

So, $x = \frac{96}{2} = 48$ mm Hg

The pressure of A remaining at time t is $P_A = P_0 - x = P_0 - 48$ mm Hg.

The time given is 5 minutes and 20 seconds. Convert this to seconds:

$t = (5 \times 60) \text{ s} + 20 \text{ s} = 300 \text{ s} + 20 \text{ s} = 320 \text{ s}$

The reaction is a first-order reaction, and the rate constant (k) is given as $5.2 \times 10^{-4} \text{ s}^{-1}$.

For a first-order reaction, the integrated rate law in terms of pressure is:

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

Substitute the known values into the equation:

$5.2 \times 10^{-4} \text{ s}^{-1} = \frac{1}{320 \text{ s}} \ln \left( \frac{P_0}{P_0 - 48} \right)$

Rearrange the equation to solve for the logarithmic term:

$\ln \left( \frac{P_0}{P_0 - 48} \right) = (5.2 \times 10^{-4} \text{ s}^{-1}) \times (320 \text{ s})$

$\ln \left( \frac{P_0}{P_0 - 48} \right) = 0.1664$

We are given that $\ln(1.181) = 0.1664$.

Therefore, we can equate the terms:

$\frac{P_0}{P_0 - 48} = 1.181$

Now, solve for $P_0$:

$P_0 = 1.181 (P_0 - 48)$

$P_0 = 1.181 P_0 - (1.181 \times 48)$

$P_0 = 1.181 P_0 - 56.688$

$56.688 = 1.181 P_0 - P_0$

$56.688 = (1.181 - 1) P_0$

$56.688 = 0.181 P_0$

$P_0 = \frac{56.688}{0.181}$

$P_0 \approx 313.193$

Rounding to the nearest integer, the initial pressure of A is 313 mm Hg.