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Question

The rate of a reaction doubles when its temperature changes from $$300\  K$$ to $$310\ K$$. Activation energy of such a reaction will be:
$$\left( R=8.314\ J\ { K }^{ -1 }{ mol }^{ -1 }\ and\ \log { 2 } =0.301 \right) $$


A
58.5 kJ mol1
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B
60.5 kJ mol1
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C
53.6 kJ mol1
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D
48.6 kJ mol1
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Solution

The correct option is D $$53.6$$ $$kJ$$ $${ mol }^{ -1 }$$
$$\log { \left( \dfrac { { k }_{ 2 } }{ { k }_{ 1 } }  \right)  } =\dfrac { { E }_{ a } }{ 2.303R } \left[ \dfrac { 1 }{ { T }_{ 1 } } -\dfrac { 1 }{ { T }_{ 2 } }  \right] $$

$$\log\left( { 2 }\right) =\dfrac { { E }_{ a } }{ 2.303\times 8.314 } \left[ \dfrac { 1 }{ 300 } -\dfrac { 1 }{ 310 }  \right] $$

$${ E }_{ a }=\dfrac { 0.301\times 2.303\times 8.314\times 300\times 310 }{ 10 } $$

   $$=53598$$ $$J$$ $${ mol }^{ -1 }$$
   $$=53.6$$ $$kJ$$ $${ mol }^{ -1 }$$

Chemistry

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