Question

For a given reaction
$H_{2}O\left(l\right)\rightleftharpoons H_{2}O\left(g\right)$ takes place in closed vessel at a constant temperature, $p_{H_{2}O}$ is the vapour pressure of $H_{2}O$. The equilibrium constant, $K_p$ is given by

• A. $K_P=p_{H_{2}O\left(g\right)}$
• B. $K_P=K_C=1$
• C. $K_p=p_{H_{2}O}^2$
• D. None of the above

Answer 1

The correct option is A $K_P=p_{H_{2}O\left(g\right)}$

Vapour pressure is the pressure exerted by vapours in thermodynamic equilibrium with its condensed phase at a given temperature in a closed system.

In the reaction,
$H_{2}O\left(l\right)\to H_{2}O\left(g\right)$
At equilibrium​,
Rate of evaporation = Rate of condensation
Hence,
vapour pressure is the equilibrium constant of the reaction​.
$K_p=p_{H_{2}O}\left(g\right)$
Thus, option (a) is correct.