g)}.$$

Assume that initially, the system contains only A with a pressure of $$P_i$$.

Define the change in pressure due to the reaction. Let “$$x$$” be the pressure (in atm or any unit) corresponding to the amount of A that has reacted at time $$t$$. Then:

The pressure of A at time $$t$$ is:

$$P_A = P_i - x.$$

Since one mole of A gives one mole of B and one mole of C, the pressures due to products are:

$$P_B = x \quad \text{and} \quad P_C = x.$$

Calculate the total pressure at time $$t$$:

$$P_t = P_A + P_B + P_C = (P_i - x) + x + x = P_i + x.$$

Thus, we can express $$x$$ in terms of the pressures:

$$x = P_t - P_i.$$

For a first order reaction, the integrated rate law using concentrations is:

$$\ln\left(\frac{[A]_0}{[A]_t}\right) = kt.$$

In the gas phase (at constant volume and temperature), pressures are proportional to concentrations. Therefore, we can write:

$$\ln\left(\frac{P_i}{P_A}\right) = kt.$$

Substitute the expression for $$P_A$$:

$$P_A = P_i - x = P_i - (P_t - P_i) = 2P_i - P_t.$$

Now the integrated rate law becomes:

$$\ln\left(\frac{P_i}{2P_i - P_t}\right) = kt.$$

Converting from the natural logarithm to a common logarithm (base 10) gives:

$$k = \frac{2.303}{t}\log\left(\frac{P_i}{2P_i - P_t}\right).$$

This is exactly Option B.

Chemical Kinetics

Continue this paper

Similar Chemical Kinetics PYQs

More practice from this question