Question

Sulfuryl chloride (So₂Cl₂) decomposes to sulphur dioxide (SO₂) and chlorine (Cl₂) by reaction in the gas phase. The following data were obtained when a sample containing 5.00×10^-2 mole of sulfuryl chloride was heated to 600 K ±1 K in a 5.00 ×10^-1 L container

Define the rate as –d[So₂Cl₂]/dt

a. Determine the value of the rate constant for the decomposition of sulfuryl chloride at 600K.

b. What is the half-life of the reaction?

c. What would be the pressure in the vessel after 0.500 h and after 12.0 h?

d. What fraction of the sulfuryl chloride remains after 20.0 h?

Short answer

The term −dtdx in the rate expression refers to the instantaneous rate of the reaction. It is the ratio of the decrease in the concentration of a reaction to the time taken for this decrease.

a. The gas phase decomposition of sulfuryl chloride at 600 K SO₂Cl₂(g) SO₂(g) + Cl₂(g) is first order in SO₂Cl₂ with a rate constant of 2.80×10^-3 min^-1.

b. Half-life is the time at which the concentration of a particular species is reduced to half equation for first order reaction is given as follows:

c. The time is 0.500 h, P1 is 4.93 and k is 0.168 / h. The values are substituted in integrated rate of first order reaction and the pressure of reactant is calculated as shown below:

$$\begin{gathered} \ln \left[P_1\right]=-kt+\ln \left[P_1\right] \\ =-\left(0.168h^{-1}\right)\left(0.500h\right)+\ln \left[4.93\right] \\ {p}_1=e^{1.51} \\ =4.53atm \end{gathered}$$

Therefore, the pressure SO₂Cl₂ at 0.500 h is 4.53 atm

The pressure of gas (SO₂ or Cl₂) is evaluated as follows:

$$\begin{gathered} P=P_1-S{O}_{2}C{l}_2 \\ =\left(4.93-4.53\right)atm \\ =0.40atm \\ P=P{}_{S{O}_{2}C{l}_2}+PS{O}_2+PC{l}_2 \\ =\left(4.53+0.40+0.40\right)atm \end{gathered}$$

Hence, the pressure of gas in the vessel at 0.500 h is 5.33 atm.

d. The time is 20.0 h and the rate constant is 0.1668/h. The values are substituted in integrated rate of first order reaction and the fraction of sulfuryl chloride is calculated as follows:

$$\begin{gathered} \ln \left[P_{S{o}_{2}C{l}_2}\right]=-\left(0.168h^{-1}\right)\left(20.0h\right)+\ln \left[P_1\right] \\ \ln \frac{\left[P_{S{o}_{2}C{l}_2}\right]}{\left[P_1\right]}=-3.36 \\ \frac{P_{S{o}_{2}C{l}_2}}{P_1}=e^{-3.36} \\ =3.47\% \end{gathered}$$

Therefore, the fraction of sulfuryl chloride remaining is 3.47 percent.

Step-by-step solution

Determine the rate constant

The rate constant, or the specific rate constant, is the proportionality constant in the equation that expresses the relationship betweenthe rate of a chemical reaction and the concentration of the reacting substances.

Determine the half-life of the reaction

Half-Life of a Chemical Reaction

1. For a zero-order reaction, the mathematical expression that can be employed to determine the half-life is: t_1/2 = [R]₀/2k.
2. For a first-order reaction, the half-life is given by: t_1/2 = 0.693/k.
3. For a second-order reaction, the formula for the half-life of the reaction is: 1/k[R]₀