JEE Advance - Chemistry (1998 - No. 4)
The rate constant of a reaction is 1.5 $$\times$$ 107 s-1 at 50oC and 4.5 $$\times$$ 107 s-1 at 100oC. Evaluate the Arrhenius parameters A and Ea.
A = 5.42 × 1010 s-1, Ea = 2.2 × 104 J mol-1
A = 2.2 × 104 s-1, Ea = 5.42 × 1010 J mol-1
A = 1.5 × 107 s-1, Ea = 4.5 × 107 J mol-1
A = 4.5 × 107 s-1, Ea = 1.5 × 107 J mol-1
A = 8.314 × 1010 s-1, Ea = 2.303 × 104 J mol-1
Explanation
According to Arrhenius equation, $$k = A{e^{ - {E_a}/RT}}$$
$$\therefore$$ $$\log k = \log A - {{{E_a}} \over {2.303RT}}$$ or, $$\log \left( {{{{k_2}} \over {{k_1}}}} \right) = {{{E_a}} \over {2.303R}}\left[ {{1 \over {{T_1}}} - {1 \over {{T_2}}}} \right]$$
Given : $$\log \left( {{{4.5 \times {{10}^7}} \over {1.5 \times {{10}^7}}}} \right) = {{{E_a}} \over {2.303 \times 8.314}}\left[ {{1 \over {323}} - {1 \over {373}}} \right]$$
or, $${E_a} = 2.2 \times {10^4}$$ J mol$$-$$1
From Arrhenius equation, $$k = A{e^{ - {E_a}/RT}}$$
or, $$\log k = \log A - {{{E_a}} \over {2.303RT}}$$
or, $$\log (4.5 \times {10^7}) = \log A - {{2.2 \times {{10}^4}} \over {2.303 \times 8.314 \times 373}}$$
or, $$A = 5.42 \times {10^{10}}$$ s$$-$$1.
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