JEE MAIN - Chemistry (2017 - 8th April Morning Slot - No. 19)
The rate of a reaction A doubles on increasing the temperature from 300 to 310 K. By how much, the temperature of reaction B should be Increased from 300 K so that rate doubles if activation energy of the reaction B is twice to that of reaction A.
9.84 K
4.92 K
2.45 K
19.67 K
Explanation
For reaction A, T1 = 300 K, T2 = 310 K, k2 = 2 k1
$$\log {{{k_2}} \over {{k_1}}} = {{{E_{{a_1}}}} \over {2.303R}}\left[ {{1 \over {{T_1}}} - {1 \over {{T_2}}}} \right]$$
$$ \therefore $$ $$\log {{2{k_1}} \over {{k_1}}} = \log 2 = {{{E_{{a_1}}}} \over {2.303R}}\left[ {{1 \over {300}} - {1 \over {310}}} \right]$$ ...(1)
For reaction B, T1 = 300 K, T2 = ?, k2 = 2k1, Ea2 = 2Ea1
$$\log {{2{k_1}} \over {{k_1}}} = \log 2 = {{2{E_{{a_1}}}} \over {2.303R}}\left[ {{1 \over {300}} - {1 \over {{T_2}}}} \right]$$
From eq. (i) and (ii), we get
$${{2{E_{{a_1}}}} \over {2.303R}}\left[ {{1 \over {300}} - {1 \over {{T_2}}}} \right] = {{{E_{{a_1}}}} \over {2.303R}}\left[ {{1 \over {300}} - {1 \over {310}}} \right]$$
$$ \Rightarrow $$ $$2\left[ {{1 \over {300}} - {1 \over {{T_2}}}} \right] = \left[ {{1 \over {300}} - {1 \over {310}}} \right]$$
$$ \Rightarrow $$ T2 = $${{{300 \times 310} \over {610}}}$$ $$ \times $$ 2 = 304.92 K
$$ \therefore $$ Increased temperature = (304.92 – 300) = 4.92 K
$$\log {{{k_2}} \over {{k_1}}} = {{{E_{{a_1}}}} \over {2.303R}}\left[ {{1 \over {{T_1}}} - {1 \over {{T_2}}}} \right]$$
$$ \therefore $$ $$\log {{2{k_1}} \over {{k_1}}} = \log 2 = {{{E_{{a_1}}}} \over {2.303R}}\left[ {{1 \over {300}} - {1 \over {310}}} \right]$$ ...(1)
For reaction B, T1 = 300 K, T2 = ?, k2 = 2k1, Ea2 = 2Ea1
$$\log {{2{k_1}} \over {{k_1}}} = \log 2 = {{2{E_{{a_1}}}} \over {2.303R}}\left[ {{1 \over {300}} - {1 \over {{T_2}}}} \right]$$
From eq. (i) and (ii), we get
$${{2{E_{{a_1}}}} \over {2.303R}}\left[ {{1 \over {300}} - {1 \over {{T_2}}}} \right] = {{{E_{{a_1}}}} \over {2.303R}}\left[ {{1 \over {300}} - {1 \over {310}}} \right]$$
$$ \Rightarrow $$ $$2\left[ {{1 \over {300}} - {1 \over {{T_2}}}} \right] = \left[ {{1 \over {300}} - {1 \over {310}}} \right]$$
$$ \Rightarrow $$ T2 = $${{{300 \times 310} \over {610}}}$$ $$ \times $$ 2 = 304.92 K
$$ \therefore $$ Increased temperature = (304.92 – 300) = 4.92 K
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