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Q3.21:

The following data were obtained during the first order thermal decomposition of (SO_{2}Cl_{2}) at a constant volume. ((SO_{2}Cl_{2}(g) \rightarrow SO_{2}(g) + Cl_{2}(g)))

Calculate the rate of the reaction when total pressure is 0.65 atm.

Solution :

Step 1: Establishing the Stoichiometric Table

The thermal decomposition reaction is: $SO_2Cl_2(g) \rightarrow SO_2(g) + Cl_2(g)$

Let $P_0$ be the initial pressure of $SO_2Cl_2$ at $t = 0$. At time $t$, let $x$ be the decrease in pressure of the reactant.

Time ($s$) $SO_2Cl_2(g)$ $SO_2(g)$ $Cl_2(g)$ Total Pressure ($P_t$)
$t=0$ $P_0$ $0$ $0$ $P_0$
$t=t$ $P_0 - x$ $x$ $x$ $P_0 + x$

From the table, at any time $t$, $x = P_t - P_0$. The partial pressure of the reactant is $P_{SO_2Cl_2} = P_0 - x = P_0 - (P_t - P_0) = 2P_0 - P_t$.

Step 2: Determining the Rate Constant ($k$)

For a first-order reaction, $k = \frac{2.303}{t} \log\left(\frac{P_0}{P_{SO_2Cl_2}}\right) = \frac{2.303}{t} \log\left(\frac{P_0}{2P_0 - P_t}\right)$.

Given: $P_0 = 0.5$ atm. At $t = 100$ s, $P_t = 0.6$ atm.

$k = \frac{2.303}{100} \log\left(\frac{0.5}{2(0.5) - 0.6}\right) = \frac{2.303}{100} \log\left(\frac{0.5}{0.4}\right)$ $k = \frac{2.303}{100} \times \log(1.25) \approx \frac{2.303 \times 0.0969}{100} = 2.23 \times 10^{-3} \, s^{-1}$.

Step 3: Calculating Partial Pressure at $P_t = 0.65$ atm

We need the rate when $P_t = 0.65$ atm. First, find the partial pressure of $SO_2Cl_2$ at this state:

$P_{SO_2Cl_2} = 2P_0 - P_t = 2(0.5) - 0.65 = 1.0 - 0.65 = 0.35$ atm.

Step 4: Calculating the Rate of Reaction

Per the first-order rate law: $\text{Rate} = k \cdot P_{SO_2Cl_2}$

Substituting the values derived:

$\text{Rate} = (2.23 \times 10^{-3} \, s^{-1}) \times (0.35 \, atm)$ $\text{Rate} = 7.805 \times 10^{-4} \, atm \cdot s^{-1}$ Decay Curve: $P_{SO_2Cl_2}$ vs Time t P

Final Solution: The rate of reaction when the total pressure is 0.65 atm is $7.81 \times 10^{-4} \, atm \cdot s^{-1}$.


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