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Question: The following data were obtained during the first order thermal decomposition of \(SO_{2}Cl_{2}\) at...

The following data were obtained during the first order thermal decomposition of SO2Cl2SO_{2}Cl_{2} at constant volume

SO2Cl2(g)→SO2(g)+Cl2(g)SO_{2}C{l_{2}}_{(g)} \rightarrow SO_{2(g)} + Cl_{2(g)}

ExperimentTime/s−1

Total pressure

/atm

1

2

0

100

0.5

0.6

What is the rate of reaction when total pressure is 0.650.65 atm?

A

0.35atms−10.35atms^{- 1}

B

2.235×10−3atms−12.235 \times 10^{- 3}atms^{- 1}

C

7.8×10−4atms−17.8 \times 10^{- 4}atms^{- 1}

D

1.55×10−4atms−11.55 \times 10^{- 4}atms^{- 1}

Answer

7.8×10−4atms−17.8 \times 10^{- 4}atms^{- 1}

Explanation

Solution

SO2Cl2→SO2+Cl2SO_{2}Cl_{2} \rightarrow SO_{2} + Cl_{2}

Initial pressure p0p_{0} 0 0

Pressure at time t p0−pp_{0} - p p p

Let initial pressure p0∝R0p_{0} \propto R_{0}

Pressure at time, Pt=p0−p+p+p=p0+pP_{t} = p_{0} - p + p + p = p_{0} + p

Pressure of reactants at time t, p0−p=2p0−pt∝Rp_{0} - p = 2p_{0} - p_{t} \propto R

k=2.303100log⁡p02p0−ptk = \frac{2.303}{100}\log\frac{p_{0}}{2p_{0} - p_{t}}

=2.303100log⁡0.52×0.5−0.6=2.303100log⁡1.25= \frac{2.303}{100}\log\frac{0.5}{2 \times 0.5 - 0.6} = \frac{2.303}{100}\log 1.25

=2.2318×10−3s−1= 2.2318 \times 10^{- 3}s^{- 1}

Pressure of SO2Cl2SO_{2}Cl_{2} at time t (pSO2Cl2)(p_{SO_{2}Cl_{2}})

=2p0−pt=2×0.50−0.65atm=0.35atms−1= 2p_{0} - p_{t} = 2 \times 0.50 - 0.65atm = 0.35atms^{- 1}

Rate at that time =k×pSO2Cl2= k \times p_{SO_{2}Cl_{2}}

=(2.2318×10−3)×(0.35)=7.8×10−4atms−1= (2.2318 \times 10^{- 3}) \times (0.35) = 7.8 \times 10^{- 4}atms^{- 1}