JEE Main 2017 · 8 Apr · Shift 1 · Chemistry Q39

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ChemistrySingle correct+4 / -1Question 39 of 90

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

Answer and worked solution

Official answer: B

For A, $\ln2=\dfrac{E_a}R\left(\dfrac1{300}-\dfrac1{310}\right)=\dfrac{E_a}R\cdot\dfrac1{9300}$ . For B, with $2E_a$ :
$\frac1{300}-\frac1T=\frac1{18600}\implies T=\frac{18600}{61}=304.92\text{ K}.$
The increase is 4.92 K.

Topics:Arrhenius equation

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JEE Main 2017 · 8 Apr · Shift 1, Chemistry question 39