Solveeit Logo
Login

Question

Question: Gaseous\({{A}_{2}}\) dissociates as, \({{A}_{2}}(g)\to 2A(g)\). Initial pressure of\({{A}_{2}}\) is ...

GaseousA2{{A}_{2}} dissociates as, A2(g)→2A(g){{A}_{2}}(g)\to 2A(g). Initial pressure ofA2{{A}_{2}} is 0.8atm. After 20 minutes the pressure is 1.1atm. Calculate the rate constant and half-life period for the reaction.

Explanation

Solution

. The base of this answer lies in the concept of the rate of second order reaction and half life for the second order reaction based on the equilibrium conditions that leads you to the correct answer.

Complete step by step answer:
- We have studied the half life and also determination of rate constant of the reaction on the basis of concentration of the reactants in our classes of physical chemistry.
- Now, let us look into how to calculate the rate constant when pressure is given.
To start with, let us consider the reaction that is taking place that is:
A2(g)→2A(g){{A}_{2}}(g)\to 2A(g)

InitialP0
At equilibriumP - x2x

Here, the initial pressure of the reactant is taken as P and that of the product will be zero because initially no product will be formed.
At the equilibrium the pressure of the product will be 2x
Thus, total pressure will be:
Pt=P−x+2x{{P}_{t}} = P - x + 2x
Pt=P+x{{P}_{t}} = P + x ………..(1)
Now, the initial pressure as per data is 0.8 atm = P
Thus, P=0.8atmP = 0.8 atm
The final pressure is, Pt=1.1atm{{P}_{t}} = 1.1 atm
Now, substituting these values in the equation (1)
1.1=0.8+x1.1 = 0.8 + x
⇒x=1.1−0.8=0.3atm\Rightarrow x = 1.1 - 0.8 = 0.3 atm
Thus, rate of a reaction at the time of 20 minutes is given by,
−d[A2]dt=xt=0.320-\dfrac{d[{{A}_{2}}]}{dt}=\dfrac{x}{t}=\dfrac{0.3}{20}
Thus,−d[A2]dt=0.015atm/min⁡\dfrac{-d[{{A}_{2}}]}{dt}=0.015atm/\min
Therefore, the rate of a reaction will be k = 0.015 atm / min.
Now, half life of reaction is related to the rate of reaction as,
T1/2=0.693k{{T}_{{}^{1}/{}_{2}}}=\dfrac{0.693}{k}
where, T1/2{{T}_{{}^{1}/{}_{2}}} is the half life
k is the rate of reaction.
Substituting the rate value, we get
T1/2=0.6930.015=1.03×10−2min⁡−1{{T}_{{}^{1}/{}_{2}}}=\dfrac{0.693}{0.015}=1.03\times {{10}^{-2}}{{\min }^{-1}}
Therefore, the correct answer is rate of a reaction is k = 0.015 atm / min and half life is 1.03×10−2min⁡−11.03\times {{10}^{-2}}{{\min }^{-1}}

Note: Note that while calculating the rate of a reaction in terms of concentration of the reactants, observe about which order the reaction is, that is whether first order or second because the formula of rate constant varies for each order of reaction.