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?


1 Answer 1


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.

As ammonium hydrosulfide is a solid, the reaction is only relevant at the surface of the crystals, and the "concentration" of ammonium hydrosulfide at the relevant areas will remain constant. Therefore same partial pressure of both ammonia and hydrogen sulfide is needed to maintain the equilibrium. Hence, at equilibrium, the partial pressure of both ammonia and hydrogen sulfide should be the same as the initial partial pressure.

  • $\begingroup$ can you help me write it mathematically ? would it be $(P/2)^2=1$ ? $\endgroup$
    – Heisenberg
    Commented Mar 19, 2015 at 10:04
  • $\begingroup$ @Sigma nah, just ${P_1 \over P_2} = 1$ or $P_1 = P_2$ $\endgroup$
    – busukxuan
    Commented Mar 19, 2015 at 10:19
  • $\begingroup$ $P_1$=$P_2$=??? $\endgroup$
    – Heisenberg
    Commented Mar 19, 2015 at 10:21
  • $\begingroup$ @Sigma $$P_1 = P_2 = 1 atm$$ $\endgroup$
    – busukxuan
    Commented Mar 19, 2015 at 10:27

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