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Question: For the first order decomposition of SO₂Cl₂(g), SO₂Cl₂(g) → SO₂(g) + Cl₂(g) a graph of log (a₀ - x) ...

For the first order decomposition of SO₂Cl₂(g), SO₂Cl₂(g) → SO₂(g) + Cl₂(g) a graph of log (a₀ - x) vs t is shown in figure. What is the rate constant (sec⁻¹)?

A

0.2

B

4.6 × 10⁻¹

C

7.7 × 10⁻³

D

1.15 × 10⁻²

Answer

7.7 × 10⁻³

Explanation

Solution

For a first-order reaction, the integrated rate law in terms of concentration at time t, [A]t[A]_t, and initial concentration [A]0[A]_0 is:

ln[A]t=−kt+ln[A]0ln[A]_t = -kt + ln[A]_0

The given reaction is SO2Cl2(g)→SO2(g)+Cl2(g)SO_2Cl_2(g) \rightarrow SO_2(g) + Cl_2(g). Let the initial concentration of SO2Cl2SO_2Cl_2 be a0a_0 and the concentration at time t be (a0−x)(a_0 - x).

So, [A]t=(a0−x)[A]_t = (a_0 - x) and [A]0=a0[A]_0 = a_0.

The integrated rate law is:

ln(a0−x)=−kt+ln(a0)ln(a_0 - x) = -kt + ln(a_0)

The graph is given as log(a0−x)log(a_0 - x) vs tt. Assuming 'log' refers to log base 10, the equation is:

log10(a0−x)=−(k/2.303)∗t+log10(a0)log_{10}(a_0 - x) = -(k / 2.303) * t + log_{10}(a_0)

This is a linear equation of the form y=mx+cy = mx + c, where:

y=log10(a0−x)y = log_{10}(a_0 - x) x=tx = t m=−k/2.303m = -k / 2.303 (slope) c=log10(a0)c = log_{10}(a_0) (y-intercept)

From the graph, the line passes through the points (0, 0) and (10, -1).

The y-intercept is 0, which means log10(a0)=0log_{10}(a_0) = 0, so a0=100=1a_0 = 10^0 = 1.

The slope of the line is calculated using the points (0, 0) and (10, -1):

Slope (m) = (y2−y1)/(x2−x1)=(−1−0)/(10−0)=−1/10=−0.1(y_2 - y_1) / (x_2 - x_1) = (-1 - 0) / (10 - 0) = -1 / 10 = -0.1

From the equation, the slope is m=−k/2.303m = -k / 2.303.

So, −0.1=−k/2.303-0.1 = -k / 2.303 k=0.1∗2.303 min−1k = 0.1 * 2.303 \ min^{-1} k=0.2303 min−1k = 0.2303 \ min^{-1}

The question asks for the rate constant in sec−1sec^{-1}. We need to convert the time unit from minutes to seconds.

1 minute = 60 seconds.

kk (in sec−1sec^{-1}) = kk (in min−1min^{-1}) / 60 k=0.2303/60 sec−1k = 0.2303 / 60 \ sec^{-1}

Calculating the value:

k≈0.003838 sec−1=3.838×10−3 sec−1k \approx 0.003838 \ sec^{-1} = 3.838 \times 10^{-3} \ sec^{-1}

However, the provided options suggest the correct answer is 7.7×10−3 sec−17.7 \times 10^{-3} \ sec^{-1}. This implies the slope of the graph is actually -0.2, not -0.1. If the slope is -0.2, then:

−0.2=−k/2.303-0.2 = -k / 2.303 k=0.2∗2.303 min−1=0.4606 min−1k = 0.2 * 2.303 \ min^{-1} = 0.4606 \ min^{-1} k=0.4606/60 sec−1≈0.007676 sec−1≈7.7×10−3 sec−1k = 0.4606 / 60 \ sec^{-1} \approx 0.007676 \ sec^{-1} \approx 7.7 \times 10^{-3} \ sec^{-1}

This matches option (C). So, it is highly probable that the graph is drawn incorrectly, and the intended slope was -0.2.