# Partial pressure and chemical equilibirum

Suppose I am given following reaction in a vessel: $$\ce{NH4HS(s) <=> NH3(g) + H2S(g)}$$ $P_\ce{NH3} = P_\ce{H2S} = \pu{1 atm}$ at equilibrium; if volume is doubled at this instant, then what would be $P_\ce{NH3}$ at new equilibrium?

I know that as soon as volume is doubled, pressure would drop to half at this instant, and $K_\mathrm{c}$ will have the same value at new equilibrium, but how to actually use this to calculate the partial pressure of ammonia at new equilibrium?

No, at new equilibrium $K_c$ would actually is the same value. This constant stands for reaction between these two compounds no matter the pressure, it is only affected by temperature.
• can you help me write it mathematically ? would it be $(P/2)^2=1$ ? – Heisenberg Mar 19 '15 at 10:04
• @Sigma nah, just ${P_1 \over P_2} = 1$ or $P_1 = P_2$ – busukxuan Mar 19 '15 at 10:19
• $P_1$=$P_2$=??? – Heisenberg Mar 19 '15 at 10:21
• @Sigma $$P_1 = P_2 = 1 atm$$ – busukxuan Mar 19 '15 at 10:27