Consider the reaction X $$\rightleftharpoons$$ Y at 300 K. If $$\Delta H^\theta$$ and K are 28.40 kJ mol$$^{-1}$$ and $$1.8 \times 10^{-7}$$ at the same temperature, then the magnitude of $$\Delta S^\theta$$ for the reaction in J K$$^{-1}$$ mol$$^{-1}$$ is _______. (Nearest integer)
(Given: R = 8.3 J K$$^{-1}$$ mol$$^{-1}$$, ln 10 = 2.3, log 3 = 0.48, log 2 = 0.30)

Step 1: Convert Given Units to SI System

• Temperature $$T=300K$$
• Enthalpy $$\triangle H^{\circ\ }=28.40kJmol^{-1}=28400\ Jmol^{-1}\ $$
• Equilibrium Constant $$K=1.8\ \times\ 10^{-7}$$
  

Step 2: Calculate the Standard Gibbs Free Energy Change $$\triangle\ G^{\circ\ }$$

The relation between Gibbs free energy and the equilibrium constant is given by:

$$\triangle\ G^{\circ\ }$$ $$=RTlnK$$

Using the relation $$\ln K=\ln10\times\ \log_{10}K$$:

$$\log_{10}K=\log_{10}\left(1.8\times\ 10^{-7}\right)=\log_{10}\left(1.8\right)-7$$

We can express $$\log_{10}\left(1.8\right)$$ using given values:

$$\log_{10}\left(1.8\right)=\log_{10}\left(\frac{18}{10}\right)=\log_{10}\left(2\times\ 3^2\right)-\log_{10}\left(10\right)$$

$$\log_{10}\left(1.8\right)=\log_{10}2+2\log_{10}3-1$$

$$\log_{10}\left(1.8\right)=0.30+2\left(0.48\right)-1=0.30+0.96-1=0.26$$

Now, substitute this back to find $$\log_{10}K:$$

$$\log_{10}K=0.26-7=-6.74$$

Next, convert to natural logarithm:

$$\ln K=2.3\times\ \left(-6.74\right)=-15.502$$

Now, solve for $$\triangle\ G^{\circ\ }:$$

$$\triangle\ G^{\circ}=-\left(8.3\ J\ K^{-1\ }mol^{-1}\right)\times\ \left(-15.502\right)=+38600\ J\ mol^{-1}$$

Step 3: Calculate the Standard Entropy Change $$\triangle\ S^{\circ\ }$$

Using thermodynamic relation:

$$\triangle\ G^{\circ\ }=\triangle\ H^{\circ\ }-T\triangle\ S^{\circ\ }$$

$$T\triangle\ S^{\circ\ }=\triangle\ H^{\circ\ }-\triangle\ G^{\circ\ }$$

$$300\times\ \triangle\ S^{\circ\ }=28400-38600\ $$

$$300\times\ \triangle\ S^{\circ\ }=-10200\ $$

$$\triangle\ S^{\circ\ }=-\frac{10200}{300}=-34\ J\ K^{-1\ }mol^{-1}$$

Therefore, the magnitude of $$\left|\triangle\ S^{\circ\ }\right|=34$$

Therefore, the correct is 34.