Increase in the pressure for the following
Equilibrium $H_{2} O (l) ⇌ H_{2} O (g)$ result in the

formation of more

H_{2} O (l)

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

formation of more

H_{2} O (g)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

increase in the boiling point of

H_{2} O (l)

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

decrease in boiling point of

H_{2} O (l)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct options are
A formation of more $H_{2} O (l)$
C increase in the boiling point of $H_{2} O (l)$

Suggest Corrections

Similar questions

H_{2} O (l) ⇌ H_{2} O (g)

H_{2} O (l) ⇌ H_{2} O (g)

H_{2} O (l)

H_{2} O (g)

H_{2} O (g) ⟶ H_{2} O (l)

Assertion (A): The enthalpy of formation of $H_{2} O_{(l)}$ is greater than that of $H_{2} O (g)$

Reason (R): Enthalpy change is negative for the condensation reaction $H_{2} O (g) \to H_{2} O (l)$

H_{2} O (l)

H_{2} O (g)

H_{2} O (g) ⟶ H_{2} O (l)

Join BYJU'S Learning Program