Question

Calculate for the equilibrium,

$\mathrm { NH } _ { 3 ( \mathrm {~g} ) } + \mathrm { H } _ { 2 } \mathrm {~S} _ { ( \mathrm { g } ) }$

If the total pressure inside the reaction vessel is $1.12$atm at $105 ^ { \circ } \mathrm { C }$ .

• A. $0.56$
• B. $1.25$
• C. $0.31$
• D. $0.63$

Answer 1

Answer: (C)

$0.31$

Answer 2

: $\mathrm { NH } _ { 3 ( \mathrm {~g} ) } + \mathrm { H } _ { 2 } \mathrm {~S} _ { ( \mathrm { g } ) }$

Initial moles 1 0 0

At equilibrium (1 – x) x x

Total gaseous moles at equilibrium = x + x = 2x

We know

But,

Partial pressure (P) = mole fraction × total pressure (P)

IInd method: Both $\mathrm { NH } _ { 3 }$ and $\mathrm { H } _ { 2 } \mathrm {~S}$ have same number of moles at equilibrium so have same mole fraction and thus equal partial pressures.

i.e.,