Q.75. Consider the reaction X Y at 300 K. If and K are 28.40 kJ mol and 1.8 10 at the same temper

Q: What is the correct answer for JEE Main Chemistry Q75?

The correct answer is 34. We are given the following information for the reaction X $\rightleftharpoons$ Y at 300 K: Temperature (T) = 300 K Standard enthalpy change ($\Delta H^0$) = 28.40 kJ mol$^{-1}$ = 28400 J mol$^{-1}$ Equilibrium constant (K) = 1.8 $\times$ 10$^{-7}$ Gas constant (R) = 8.3 J K$^{-1}$ mol$^{-1}$ We need to find the standard entropy change ($\Delta S^0$). The relationship between $\Delta G^0$, $\Delt

Q: Q.75. Consider the reaction X Y at 300 K. If and K are 28.40 kJ mol and 1.8 10 at the same temperature, then the magnitude of for the reaction in J K mol is ______.(Nearest Integer). (Given : R =…

The correct answer is 34. We are given the following information for the reaction X $\rightleftharpoons$ Y at 300 K: Temperature (T) = 300 K Standard enthalpy change ($\Delta H^0$) = 28.40 kJ mol$^{-1}$ = 28400 J mol$^{-1}$ Equilibrium constant (K) = 1.8 $\times$ 10$^{-7}$ Gas constant (R) = 8.3 J K$^{-1}$ mol$^{-1}$ We need to find the standard entropy change ($\Delta S^0$). The relationship between $\Delta G^0$, $\Delt

Question

Q.75. Consider the reaction X  Y at 300 K. If  and K are 28.40 kJ mol and 1.8  10 at the same temperature, then the magnitude of  for the reaction in J K mol is ______.(Nearest Integer). (Given : R = 8.3 J K mol, ln10 = 2.3, log 3 = 0.48, log 2 = 0.30

FirstInTest · Accepted Answer

We are given the following information for the reaction X $ightleftharpoons$ Y at 300 K:
Temperature (T) = 300 K
Standard enthalpy change ($\Delta H^0$) = 28.40 kJ mol$^{-1}$ = 28400 J mol$^{-1}$
Equilibrium constant (K) = 1.8 $	imes$ 10$^{-7}$
Gas constant (R) = 8.3 J K$^{-1}$ mol$^{-1}$

We need to find the standard entropy change ($\Delta S^0$).

The relationship between $\Delta G^0$, $\Delta H^0$, and $\Delta S^0$ is:
$\Delta G^0 = \Delta H^0 - T\Delta S^0$

Also, $\Delta G^0$ is related to the equilibrium constant K by:
$\Delta G^0 = -RT \ln K$

Equating the two expressions for $\Delta G^0$:
$-RT \ln K = \Delta H^0 - T\Delta S^0$

Rearranging to solve for $\Delta S^0$:
$T\Delta S^0 = \Delta H^0 + RT \ln K$
$\Delta S^0 = \frac{\Delta H^0}{T} + R \ln K$

Now, substitute the given values:
$\Delta S^0 = \frac{28400 	ext{ J mol}^{-1}}{300 	ext{ K}} + (8.3 	ext{ J K}^{-1} 	ext{ mol}^{-1}) \ln (1.8 	imes 10^{-7})$

First term: $\frac{28400}{300} = 94.666... 	ext{ J K}^{-1} 	ext{ mol}^{-1}$

Second term: $8.3 \ln (1.8 	imes 10^{-7})$
We can write $\ln (1.8 	imes 10^{-7}) = \ln (1.8) + \ln (10^{-7}) = \ln (1.8) - 7 \ln (10)$
Given $\ln 10 = 2.3$. So, $-7 \ln 10 = -7 	imes 2.3 = -16.1$
To find $\ln (1.8)$, we can use $\ln (1.8) = \ln (\frac{18}{10}) = \ln (\frac{9}{5}) = \ln (9) - \ln (5) = \ln (3^2) - \ln (5) = 2 \ln 3 - \ln 5$.
Given $\log 3 = 0.48$, so $\ln 3 = 2.3 	imes 0.48 = 1.104$.
Given $\log 2 = 0.30$, so $\ln 2 = 2.3 	imes 0.30 = 0.69$.
$\ln 5 = \ln (10/2) = \ln 10 - \ln 2 = 2.3 - 0.69 = 1.61$.
So, $\ln (1.8) = 2 	imes 1.104 - 1.61 = 2.208 - 1.61 = 0.598$.

So, $\ln (1.8 	imes 10^{-7}) = 0.598 - 16.1 = -15.502$

Now, $R \ln K = 8.3 	imes (-15.502) = -128.6666 	ext{ J K}^{-1} 	ext{ mol}^{-1}$

$\Delta S^0 = 94.666... - 128.6666... = -34.00 	ext{ J K}^{-1} 	ext{ mol}^{-1}$

The magnitude of $\Delta S^0$ is 34.00 J K$^{-1}$ mol$^{-1}$.
Rounding to the nearest integer, the magnitude is 34.

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