JEE Main 2022 · 29 Jun · Shift 1 · Chemistry Q85

← JEE Main 2022 · 29 Jun · Shift 1

ChemistryInteger+4 / -1Question 85 of 90

The activation energy of one of the reactions in a biochemical process is 532611 J mol $^{-1}$ . When the temperature falls from 310 K to 300 K, the change in rate constant observed is $k_{300}=x\times10^{-3}k_{310}$ . The value of $x$ is ______.

[Given : ln10 = 2.3; R = 8.3 J K $^{-1}$ mol $^{-1}$ ]

Your answer

Answer and worked solution

Official answer: 1

Arrhenius. $\ln\dfrac{k_{310}}{k_{300}}=\dfrac{532611}{8.3}\left(\dfrac1{300}-\dfrac1{310}\right)=6.9=3\ln10$ .

So $k_{300}=10^{-3}k_{310}$ , $x=1$ .

Answer: 1

Topics:Arrhenius equation

← Previous question Next question →

JEE Main 2022 · 29 Jun · Shift 1, Chemistry question 85