Q.57. Given below are two statements : R = 8.314 J K⁻¹ mol⁻¹ and 1 cal = 4.2 J Statement I: When Ea

Q: What is the correct answer for JEE Main Chemistry Q57?

The correct answer is C. Statement I: For a reaction, $\ln\frac{K_2}{K_1} = \frac{E_a}{R} (\frac{T_2 - T_1}{T_1 T_2})$. Given $E_a = 12.6 \, \mathrm{kcal/mol} = 12.6 \times 4.2 \times 10^3 \, \mathrm{J/mol}$. $R = 8.314 \, \mathrm{J K^{-1} mol^{-1}}$. $T_1 = 298 \, \mathrm{K}$, $T_2 = 308 \, \mathrm{K}$. Substituting these values, $\ln\frac{K_2}{K_1} = \frac{12.6 \times 10^3 \times 4.2}{8.314} (\frac{10}{298 \times 308})

Q: Q.57. Given below are two statements : R = 8.314 J K⁻¹ mol⁻¹ and 1 cal = 4.2 J Statement I: When Ea = 12.6 kcal/mol, the room temperature rate constant is doubled by a 10°C increase in temperature…

The correct answer is C. Statement I: For a reaction, $\ln\frac{K_2}{K_1} = \frac{E_a}{R} (\frac{T_2 - T_1}{T_1 T_2})$. Given $E_a = 12.6 \, \mathrm{kcal/mol} = 12.6 \times 4.2 \times 10^3 \, \mathrm{J/mol}$. $R = 8.314 \, \mathrm{J K^{-1} mol^{-1}}$. $T_1 = 298 \, \mathrm{K}$, $T_2 = 308 \, \mathrm{K}$. Substituting these values, $\ln\frac{K_2}{K_1} = \frac{12.6 \times 10^3 \times 4.2}{8.314} (\frac{10}{298 \times 308})

Q: What are the options for this JEE Main Chemistry question?

A. Both Statement I and Statement II are true · B. Both Statement I and Statement II are false · C. Statement I is true but Statement II are false · D. Statement I is false but Statement II are true

Question

Q.57. Given below are two statements : R = 8.314 J K⁻¹ mol⁻¹ and 1 cal = 4.2 J Statement I: When Ea = 12.6 kcal/mol, the room temperature rate constant is doubled by a 10°C increase in temperature (298 K to 308 K) Statement II: For a first order reaction A → B, [Figure: Graph of t1/2 vs [A] showing t1/2 is constant.] Here [A] is the initial concentration of A and t₁/₂ is half-life of reaction. In the light of the above statements, choose the correct answer from the option given below :

FirstInTest · Accepted Answer

Statement I: For a reaction, $\ln\frac{K_2}{K_1} = \frac{E_a}{R} (\frac{T_2 - T_1}{T_1 T_2})$. Given $E_a = 12.6 \, \mathrm{kcal/mol} = 12.6 	imes 4.2 	imes 10^3 \, \mathrm{J/mol}$. $R = 8.314 \, \mathrm{J K^{-1} mol^{-1}}$. $T_1 = 298 \, \mathrm{K}$, $T_2 = 308 \, \mathrm{K}$. Substituting these values, $\ln\frac{K_2}{K_1} = \frac{12.6 	imes 10^3 	imes 4.2}{8.314} (\frac{10}{298 	imes 308}) \approx 0.693$. Since $\ln 2 \approx 0.693$, $\frac{K_2}{K_1} \approx 2$. This means the rate constant doubles, so Statement I is true. Statement II: For a first order reaction, the half-life $t_{1/2} = \frac{\ln 2}{K}$. This equation shows that $t_{1/2}$ is independent of the initial concentration [A]. The graph correctly depicts this, but the statement implies $t_{1/2}$ is proportional to [A], which is incorrect. Therefore, Statement II is false.

Answer

Both Statement I and Statement II are true

Answer

Both Statement I and Statement II are false

Answer

Statement I is true but Statement II are false

Answer

Statement I is false but Statement II are true

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