Question

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

• 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

The correct option is A formation of more $H_{2}O\left(s\right)$

$H_{2}O\left(s\right)\rightleftharpoons H_{2}O\left(l\right)$

According to Le-chatelier principle decrease in pressure indicates more fromation of $H_{2}O\left(s\right)$ .