For the equilibrium reaction- H-2-O-l- rightleftharpoons H-2-O-g- what happens- pressure is applied -More water evaporatesThe boiling point of water is increasedNo effect on boiling pointNone of the above

Question

Toppr Admin · Accepted Answer

The equilibrium reaction is H2O-l-x21CC-H2O-g- when pressure is applied- the equilibrium will shift to left as the value of -x394-n is positive- Hence- for boiling to occur more temperature is required- So- it will increase the boiling point-

For the equilibrium reaction, $H_{2} O (l) ⇌ H_{2} O (g),$ what happens, if pressure is applied :
More water evaporates
The boiling point of water is increased
No effect on boiling point
None of the above

The equilibrium reaction is $H_{2} O (l) ⇌ H_{2} O (g),$ when pressure is applied, the equilibrium will shift to left as the value of $Δ n$ is positive. Hence, for boiling to occur more temperature is required. So, it will increase the boiling point.

For the equilibrium reaction, H2O(l)⇌H2O(g), what happens, if pressure is applied :More water evaporatesThe boiling point of water is increasedNo effect on boiling pointNone of the above

The equilibrium reaction is H2O(l)⇌H2O(g), when pressure is applied, the equilibrium will shift to left as the value of Δn is positive. Hence, for boiling to occur more temperature is required. So, it will increase the boiling point.

For the equilibrium reaction, $H_{2} O (l) ⇌ H_{2} O (g),$ what happens, if pressure is applied :
More water evaporates
The boiling point of water is increased
No effect on boiling point
None of the above

The equilibrium reaction is $H_{2} O (l) ⇌ H_{2} O (g),$ when pressure is applied, the equilibrium will shift to left as the value of $Δ n$ is positive. Hence, for boiling to occur more temperature is required. So, it will increase the boiling point.