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Question: The time required for \(10\% \) completion of a first order reaction at \(298{\text{ K}}\) is equal ...

The time required for 10%10\% completion of a first order reaction at 298 K298{\text{ K}} is equal to that required for its 25%25\% completion at 308 K308{\text{ K}}. If the pre-exponential factor for the reaction is 3.56×109 sec13.56 \times {10^9}{\text{ se}}{{\text{c}}^{ - 1}}, calculate its rate constant at 318 K318{\text{ K}} and also the energy of activation.

Explanation

Solution

To solve this question we must know the equation for the rate constant of first order reaction. Using the equation calculate the rate constant at 10%10\% completion and 25%25\% completion. The minimum amount of energy that the reacting species must possess to undergo a specific reaction is known as the activation energy. The relation between the temperature, rate constant and activation energy is given by the Arrhenius equation.

Formulae Used:
1. k=2.303tlog[a]0[a]k = \dfrac{{2.303}}{t}\log \dfrac{{{{\left[ a \right]}^0}}}{{\left[ a \right]}}
2. logk2k1=Ea2.303×R(1T11T2)\log \dfrac{{{k_2}}}{{{k_1}}} = \dfrac{{{E_a}}}{{2.303 \times R}}\left( {\dfrac{1}{{{T_1}}} - \dfrac{1}{{{T_2}}}} \right)
3. logk=logAEa2.303×R×T\log k = \log A - \dfrac{{{E_a}}}{{2.303 \times R \times T}}

Complete step by step solution:
We know the equation for the rate constant of a first order reaction is,
k=2.303tlog[a]0[a]k = \dfrac{{2.303}}{t}\log \dfrac{{{{\left[ a \right]}^0}}}{{\left[ a \right]}}
where kk is the rate constant of a first order reaction,
tt is time,
[a]0{\left[ a \right]^0} is the initial concentration of the reactant,
[a]\left[ a \right] is the final concentration of the reactant.
The reaction is 10%10\% complete. Thus, the initial concentration is 100{\text{100}} and the final concentration is 10010=90100 - 10 = 90. Thus,
k298 K=2.303tlog10090{k_{{\text{298 K}}}} = \dfrac{{2.303}}{t}\log \dfrac{{100}}{{90}}
The reaction is 25%25\% complete. Thus, the initial concentration is 100{\text{100}} and the final concentration is 10025=75100 - 25 = 75. Thus,
k308 K=2.303tlog10075{k_{{\text{308 K}}}} = \dfrac{{2.303}}{t}\log \dfrac{{100}}{{75}}
Thus,
k308 Kk298 K=2.303tlog100752.303tlog10090\dfrac{{{k_{{\text{308 K}}}}}}{{{k_{{\text{298 K}}}}}} = \dfrac{{\dfrac{{2.303}}{t}\log \dfrac{{100}}{{75}}}}{{\dfrac{{2.303}}{t}\log \dfrac{{100}}{{90}}}}
k308 Kk298 K=2.73\dfrac{{{k_{{\text{308 K}}}}}}{{{k_{{\text{298 K}}}}}} = 2.73
We know the Arrhenius equation,
logk2k1=Ea2.303×R(1T11T2)\log \dfrac{{{k_2}}}{{{k_1}}} = \dfrac{{{E_a}}}{{2.303 \times R}}\left( {\dfrac{1}{{{T_1}}} - \dfrac{1}{{{T_2}}}} \right)
where k2 and k1{k_2}{\text{ and }}{k_1} are the constants for the reaction,
Ea{E_a} is the energy of activation,
RR is the universal gas constant,
T1 and T2{T_1}{\text{ and }}{T_2} are the temperatures.
Thus, for the given first order reaction, the Arrhenius equation is,
logk308 Kk298 K=Ea2.303×R(1T11T2)\log \dfrac{{{k_{308{\text{ K}}}}}}{{{k_{298{\text{ K}}}}}} = \dfrac{{{E_a}}}{{2.303 \times R}}\left( {\dfrac{1}{{{T_1}}} - \dfrac{1}{{{T_2}}}} \right)
Substitute k308 Kk298 K=2.73\dfrac{{{k_{308{\text{ K}}}}}}{{{k_{298{\text{ K}}}}}} = 2.73, 8.314 J K1 mol18.314{\text{ J }}{{\text{K}}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}} for the universal gas constant, T1=298 K{T_1} = 298{\text{ K}} and T2=308 K{T_2} = 308{\text{ K}}. Thus,
log2.73=Ea2.303×8.314 J K1 mol1(1298 K1308 K)\log 2.73 = \dfrac{{{E_a}}}{{2.303 \times 8.314{\text{ J }}{{\text{K}}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}}}}\left( {\dfrac{1}{{298{\text{ K}}}} - \dfrac{1}{{308{\text{ K}}}}} \right)
Ea=76623 J mol1{E_a} = 76623{\text{ J mo}}{{\text{l}}^{ - 1}}
Thus, the energy of activation is 76623 J mol176623{\text{ J mo}}{{\text{l}}^{ - 1}}.
Calculate the rate constant at 318 K318{\text{ K}} using the Arrhenius equation,
logk=logAEa2.303×R×T\log k = \log A - \dfrac{{{E_a}}}{{2.303 \times R \times T}}
where kk is the rate constant of a first order reaction,
AA is the pre-exponential factor,
Ea{E_a} is the energy of activation,
RR is the universal gas constant,
TT is the temperature.
Thus,
logk=log(3.56×109 sec1)76623 J mol12.303×8.314 J K1 mol1×318 K\log k = \log \left( {3.56 \times {{10}^9}{\text{ se}}{{\text{c}}^{ - 1}}} \right) - \dfrac{{76623{\text{ J mo}}{{\text{l}}^{ - 1}}}}{{2.303 \times 8.314{\text{ J }}{{\text{K}}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}} \times 318{\text{ K}}}}
logk=3.0328\log k = - 3.0328
k=9.27×104 sec1k = 9.27 \times {10^{ - 4}}{\text{ se}}{{\text{c}}^{ - 1}}

**Thus, the rate constant at 318 K318{\text{ K}} is 9.27×104 sec19.27 \times {10^{ - 4}}{\text{ se}}{{\text{c}}^{ - 1}}.

Note: **
The unit of rate constant for first order reaction is sec1{\text{se}}{{\text{c}}^{ - 1}}. The units do not contain concentration terms. Thus, we can say that the rate constant of a first order reaction is independent of the concentration of the reactant. The rate of any chemical reaction is inversely proportional to the activation energy. Higher the activation energy, slower is the chemical reaction.