Solveeit Logo
Login

Question

Question: The rate coefficient \((k)\) for a particular reaction is \(1.3 \times {10^{ - 4}}{M^{ - 1}}{s^{ - 1...

The rate coefficient (k)(k) for a particular reaction is 1.3×10−4M−1s−11.3 \times {10^{ - 4}}{M^{ - 1}}{s^{ - 1}} at 100∘C{100^ \circ }Cand 1.3×10−3M−1s−11.3 \times {10^{ - 3}}{M^{ - 1}}{s^{ - 1}} at 150∘C{150^ \circ }C . What is the energy of activation (in kJ)({\text{in }}kJ) for this reaction?
(A)1616
(B)6060
(C)9999
(D)132132

Explanation

Solution

Hint : The Arrhenius equation gives the relation between the rate of the reaction and temperature for many chemical and physical reactions. The equation is temperature dependent. The Arrhenius equation helps in solving problems related to the rate of reactions. The Arrhenius equation helps in calculating the activation energy if rate constants are known.
Formula used:
ln⁡k2k1=EaR(1T1−1T2)\ln \dfrac{{{k_2}}}{{{k_1}}} = \dfrac{{{E_a}}}{R}\left( {\dfrac{1}{{{T_1}}} - \dfrac{1}{{{T_2}}}} \right)

Complete Step By Step Answer:
Given in question that:
k1=1.3×10−4M−1s−1{k_1} = 1.3 \times {10^{ - 4}}{M^{ - 1}}{s^{ - 1}}
k2=1.3×10−3M−1s−1{k_2} = 1.3 \times {10^{ - 3}}{M^{ - 1}}{s^{ - 1}}
T1=373K{T_1} = 373K
T2=423K{T_2} = 423K
We know that value of RRat standard atmospheric pressure is R=8.314Jmol−1K−1R = 8.314Jmo{l^{ - 1}}{K^{ - 1}}
By substituting all these values in Arrhenius equation we get:
ln⁡k2k1=EaR(1T1−1T2)\ln \dfrac{{{k_2}}}{{{k_1}}} = \dfrac{{{E_a}}}{R}\left( {\dfrac{1}{{{T_1}}} - \dfrac{1}{{{T_2}}}} \right)
ln⁡1.3×10−31.3×10−4=Ea8.314(1373−1423)\ln \dfrac{{1.3 \times {{10}^{ - 3}}}}{{1.3 \times {{10}^{ - 4}}}} = \dfrac{{{E_a}}}{{8.314}}\left( {\dfrac{1}{{373}} - \dfrac{1}{{423}}} \right)
On solving further we get,
Ea=60kJ{E_a} = 60kJ

Therefore, option B is correct.

Note :
In the Arrhenius equation, kkis the rate constant, AAis called the Arrhenius constant which is also called a frequency factor, Ea{E_a}is the activation energy and RRis the universal gas constant. We also have TTwhich is the temperature in kelvin. The Arrhenius constant takes into account the frequency of collisions with the correct orientations for the reaction to occur. The activation energy can be calculated from the value of rate constants at two different temperatures. An increase in temperature, generally speeds up the reaction because the rate constant increases according to the Arrhenius equation.