JEE MAIN - Chemistry (2019 - 10th April Evening Slot - No. 15)
For the reaction of H2 with I2, the rate constant is 2.5 × 10–4 dm3
mol–1s–1
at 327°C and 1.0 dm3
mol–1
at
527°C. The activation energy for the reaction, in kJ mole–1
is : (R = 8.314 JK–1
mol–1
)
59
166
72
150
Explanation
From Arrhenius equation, we get
$$\log {{{K_2}} \over {{K_1}}} = {{{E_a}} \over {2.303R}}\left( {{1 \over {{T_1}}} - {1 \over {{T_2}}}} \right)$$
$$ \Rightarrow $$ $$\log {1 \over {2.5 \times {{10}^{ - 4}}}} = {{{E_a}} \over {2.303 \times 8.314}}\left( {{1 \over {600}} - {1 \over {800}}} \right)$$
$$ \Rightarrow $$ 3.6 = $${{{E_a}} \over {2.303 \times 8.314}} \times {{200} \over {600 \times 800}}$$
$$ \Rightarrow $$ E$$a$$ = 166 kJ/mol
$$\log {{{K_2}} \over {{K_1}}} = {{{E_a}} \over {2.303R}}\left( {{1 \over {{T_1}}} - {1 \over {{T_2}}}} \right)$$
$$ \Rightarrow $$ $$\log {1 \over {2.5 \times {{10}^{ - 4}}}} = {{{E_a}} \over {2.303 \times 8.314}}\left( {{1 \over {600}} - {1 \over {800}}} \right)$$
$$ \Rightarrow $$ 3.6 = $${{{E_a}} \over {2.303 \times 8.314}} \times {{200} \over {600 \times 800}}$$
$$ \Rightarrow $$ E$$a$$ = 166 kJ/mol
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