Question

The vapour pressure of liquid methanol at ${50}^{0}C$ is $55.5kPa$. Which one is correct for equilibrium reaction attained in a vessel of 5 litres at ${50}^{0}C$ for
$C{H}_{3}OH\left(l\right)\rightleftharpoons C{H}_{3}OH\left(g\right)$?

• A. $K_p=55.5kPa$
• B. $K_c=2.067mol{litre}^{-1}$
• C. $K^0=0.555$
• D. $K^0=0.555kPa$

Answer 1

Answer: (A), (C)

The correct options are
A $K_p=55.5kPa$
C $K^0=0.555$

The vapour pressure of liquid methanol at $50$ is $55.5kPa$. The statements A, B and C are correct for equilibrium reaction attained in a vessel of 5 litre at $50$ for
$C{H}_{3}OH\left(l\right)\rightleftharpoons C{H}_{3}OH\left(g\right)$

(A) $K_p=55.5kPa$.

The equilibrium constant in terms of partial pressures is equal to the partial pressure of methanol which is 55.5 kPa. Note: The active masses of solids and liquids are assumed unity.

(B) $K_c=2.067mol{litre}^{-1}$

$K_p=55.5kPa=0.555atm$

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

$R=0.08206Latm/mol/K$

$0.555=K_c\times (0.08206\times 323{)}^1$

$K_c=0.021mol{litre}^{-1}$

(C) $K^0=0.555$. It is the ratio of the partial pressure of methanol to the partial pressure of methanol in its standard state. It is $\frac{55.5}{100}=0.555$

Equilibrium constant expressed in terms of relative activity (i.e., the activity of substance relative to its activity in standard state) has no unit

This is because relative activity is the ratio of two quantities having same units. Hence, relative activity is dimensionless.