Question

Decrease in the pressure for the following equilibria: $H_{2}O\left(s\right)\rightleftharpoons H_{2}O\left(l\right)$ result in:

• A. Formation of more $H_{2}O\left(s\right)$
• B. Formation of more $H_{2}O\left(l\right)$
• C. increase in melting point of $H_{2}O\left(s\right)$
• D. decrease in melting point of $H_{2}O\left(s\right)$

Answer 1

Answer: (A), (C)

The correct options are
A Formation of more $H_{2}O\left(s\right)$
C increase in melting point of $H_{2}O\left(s\right)$

When the pressure of the equilibrium $H_{2}O\left(s\right)\rightleftharpoons H_{2}O\left(l\right)$ is decreased, the equilibrium will shift in the backward direction so as to form more $H_{2}O\left(s\right)$. This will nullify the effect of decreased pressure.
Also the melting point of $H_{2}O\left(s\right)$ will increase as more temperature is required to shift the equilibrium towards right.