Question

3. The following data of decomposition reaction of thionyl chloride (SO2Cl2) were collected at a certain temperature and the concentration of SO2Cl2 were monitored as shown in the table.
SO2Cl2 (g) SO2 (g) + Cl2 (g)



Time (min) Conc. of SO2Cl2 (mol/L)

0 0.1000

100 0.0876

200 0.0768

300 0.0673

400 0.0590

500 0.0517

600 0.0453

700 0.0397

800 0.0348

900 0.0305

1000 0.0267

1100 0.0234



a) Determine graphically whether the kinetics of the reaction is zero order, first order or second order with respect to SO2Cl2 and then write the rate equation.

b) Determine the rate constant (k) of the reaction.

c) Determine the half-life (t½) for the reaction.

d) What will be the concentration of SO2Cl2 left in the reaction mixture at 1500 minutes?

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Answer 1

Answer:

a) First-order.

b) 0.013 min⁻¹

c) 53.3 min.

d) 0.0142M

Explanation:

Hello,

In this case, on the attached document, we can notice the corresponding plot for each possible order of reaction. Thus, we should remember that in zeroth-order we plot the concentration of the reactant (SO2Cl2 ) versus the time, in first-order the natural logarithm of the concentration of the reactant (SO2Cl2 ) versus the time and in second-order reactions the inverse of the concentration of the reactant (SO2Cl2 ) versus the time.

a) In such a way, we realize the best fit is exhibited by the first-order model which shows a straight line (R=1) which has a slope of -0.0013 and an intercept of -2.3025 (natural logarithm of 0.1 which corresponds to the initial concentration). Therefore, the reaction has a first-order kinetics.

b) Since the slope is -0.0013 (take two random values), the rate constant is 0.013 min⁻¹:

$m=\frac{ln(0.0768)-ln(0.0876)}{200min-100min} =-0.0013min^{-1}$

c) Half life for first-order kinetics is computed by:

$t_{1/2}=\frac{ln(2)}{k}=\frac{ln(2)}{0.013min^{-1}} =53.3min$

d) Here, we compute the concentration via the integrated rate law once 1500 minutes have passed:

$C=C_0exp(-kt)=0.1Mexp(-0.013min^{-1}*1500min)\\\\C=0.0142M$

Best regards.