Advertisements
Advertisements
Question
The following data were obtained during the first order thermal decomposition of SO2Cl2 at a constant volume :
SO2Cl2 (g) → SO2 (g) + Cl2 (g)
| Experiment | Time/s–1 | Total pressure/atm |
| 1 | 0 | 0.4 |
| 2 | 100 | 0.7 |
Calculate the rate constant.
(Given : log 4 = 0.6021, log 2 = 0.3010)
Advertisements
Solution 1
The given reaction is as follows:
SO2Cl2 (g) → SO2(g)+Cl2(g)
Also given:
| Experiment | Time/s–1 | Total pressure/atm |
| 1 | 0 | 0.4 |
| 2 | 100 | 0.7 |
The rate constant k can be calculated as follows:
`k=2.303/tlog `
when t= 100s, k = `2.303/(100s)log`
`k = 2.303/(100s)log`
`k=2.303/(100s) log 4`
`k = 2.303/(100s)xx0.6021`
k = 1.39 x 10-2s-1
Therefore, the rate constant is 1.39 x 10-2s-1
Solution 2
The thermal decomposition of SO2Cl2 at a constant volume is represented by the following equation:
SO2Cl2→SO2(g) + Cl2(g)
At t =0 P0 0 0
At t=t P0 −p p p
After time t, total pressure is given as:
Pt =(P0−p) + p + p
Pt = P0 + p
This, on rearrangement, gives:
p = Pt − P0
∴ P0−p=P0−(Pt−P0)= 2P0 − Pt
For the first-order reaction, we have:
`k=2.303/tlog P_0/(P_0 - p)`
`=2.303/tlog P_0/(P_0 - p)`
t = 100 s
`k = 2.303/100log 0.42/(0.4 - 0.7) = 1.386 xx 10^(-2)s-1`
APPEARS IN
RELATED QUESTIONS
A reaction is second order in A and first order in B.
(i) Write the differential rate equation.
(ii) How is the rate affected on increasing the concentration of A three times?
(iii) How is the rate affected when the concentrations of both A and B are doubled?
The conversion of molecules X to Y follows second order kinetics. If concentration of X is increased to three times how will it affect the rate of formation of Y?
Rate of reaction for the combustion of propane is equal to:
\[\ce{C3H8_{(g)} + 5O2_{(g)} -> 3CO2_{(g)} + 4H2O_{(g)}}\]
Consider a first order gas phase decomposition reaction given below :
\[\ce{A(g) -> B(g) + C(g)}\]
The initial pressure of the system before decomposition of A was pi. After lapse of time ‘t’, total pressure of the system increased by x units and became ‘pt’ The rate constant k for the reaction is given as ______.
Compounds ‘A’ and ‘B’ react according to the following chemical equation.
\[\ce{A(g) + 2B(g) -> 2C(g)}\]
Concentration of either ‘A’ or ‘B’ were changed keeping the concentrations of one of the reactants constant and rates were measured as a function of initial concentration. Following results were obtained. Choose the correct option for the rate equations for this reaction.
| Experiment | Initial concentration of [A]/mol L–¹ |
Initial concentration of [B]/mol L–¹ |
Initial rate of formation of [C]/mol L–¹ s–¹ |
| 1. | 0.30 | 0.30 | 0.10 |
| 2. | 0.30 | 0.60 | 0.40 |
| 3. | 0.60 | 0.30 | 0.20 |
For a complex reaction:
(i) order of overall reaction is same as molecularity of the slowest step.
(ii) order of overall reaction is less than the molecularity of the slowest step.
(iii) order of overall reaction is greater than molecularity of the slowest step.
(iv) molecularity of the slowest step is never zero or non interger.
Assertion: The enthalpy of reaction remains constant in the presence of a catalyst.
Reason: A catalyst participating in the reaction, forms different activated complex and lowers down the activation energy but the difference in energy of reactant and product remains the same.
The role of a catalyst is to change
For a reaction A + B → products, the rate law is given by: r = `K[A]^(1/2)`. What is the order of reaction?
If the 0.05 molar solution of m+ is replaced by a 0.0025 molar m+ solution, then the magnitude of the cell potential would be
