QJEEAdv16Ch0e56: JEE Advanced 2016 Chemistry - Class 11 JEEAdv16Ch0e56 Solution: Paragraph Thermal decomposition of gaseous X 2 to gaseous X at…

On decreasing total pressure, the reaction will move in the forward direction where number of gaseous molecules is less.

Hence, option (A) is correct.

At the start of reaction Q = 0. From $$\mathrm{\Delta }$$G = $$\mathrm{\Delta }$$G^o + RT lnQ, we have that for Q = 0 at the start of reaction $$\mathrm{\Delta }$$_rG is negative, this causes dissociation of X₂ to take place spontaneously.

Hence, option (B) is correct.

If b_equilibrium = 0.7, then value of equilibrium constant is

$$K_p=\frac{8{\beta }_{eq}^2}{4-{\beta }_{eq}^2}=\frac{8{(0.7)}^2}{7-{(0.7)}^2}$$

The value of K_p is greater than 1, which is not possible as given that $$\mathrm{\Delta }$$G^o > 0 for the reaction.

Hence, option (C) is incorrect.

As $$\mathrm{\Delta }$$G^o > 0 and $$\mathrm{\Delta }$$G^o = $$-$$RT ln(K_p)

$$\mathrm{\Delta }$$G^o > 1, so K_p should be less than 1.

We know that

$$K_p=K_c(RT{)}^{\mathrm{\Delta }{n}_g}$$

$$K_c=\frac{K_p}{{(RT)}^{\mathrm{\Delta }{n}_g}}$$

Therefore, K_c < K_p . As K_p is less than 1, so K_c is also less than 1.

Hence, option (D) is correct.