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- 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:

${{A}_{2}}(g)\to 2A(g)$

_{Initial} | _{P} | _{0} |

_{At equilibrium } | _{P - x} | _{2x} |

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:

\[{{P}_{t}} = P - x + 2x\]

\[{{P}_{t}} = P + x\] ………..(1)

Now, the initial pressure as per data is 0.8 atm = P

Thus, $P = 0.8 atm$

The final pressure is, ${{P}_{t}} = 1.1 atm$

Now, substituting these values in the equation (1)

\[1.1 = 0.8 + x\]

\[\Rightarrow x = 1.1 - 0.8 = 0.3 atm\]

Thus, rate of a reaction at the time of 20 minutes is given by,

\[-\dfrac{d[{{A}_{2}}]}{dt}=\dfrac{x}{t}=\dfrac{0.3}{20}\]

Thus,$\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,

\[{{T}_{{}^{1}/{}_{2}}}=\dfrac{0.693}{k}\]

where, \[{{T}_{{}^{1}/{}_{2}}}\] is the half life

k is the rate of reaction.

Substituting the rate value, we get

\[{{T}_{{}^{1}/{}_{2}}}=\dfrac{0.693}{0.015}=1.03\times {{10}^{-2}}{{\min }^{-1}}\]

Therefore, the correct answer is rate of a reaction is k = 0.015 atm / min and half life is \[1.03\times {{10}^{-2}}{{\min }^{-1}}\]

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