NH4I (s) <==> NH3(g) + HI(g) A mass of 5 g of NH4I is sealed in a 2.00L flask and heated to 673K. If 2.56g NH4I(s) remain unreacted when the system has reached equilibrium, what is the equilibrium constant (Kp) for the reaction? (R = 0.0821)

The value of ##K_P## is 0.2.

1. Calculate mass of NH₄I reacted.

You started with 5 g of NH₄I and ended up with 2.56 g at equilibrium. So

Mass of NH₄I reacted = 5 g – 2.56 g = 2.4 g (1 significant figure + 1 guard digit.

2. Calculate moles of NH₃ and HI formed.

NH₄I(s) ⇌ NH₃(g) + HI(g)

Moles of NH₃ formed = ##2.4″ g NH”_4″I” × (“1 mol NH”_4″I”)/(“144.94 g NH”_4″I”) × (“1 mol NH”_3)/(“1 mol NH”_4″I”) = “0.017 mol NH”_3##

Moles of HI formed = ##”0.017 mol NH”_3 × “1 mol HI”/(“1 mol NH”_3) = “0.017 mol HI”##

3. Calculate partial pressures of NH₃ and of HI.

##PV = nRT##

##P_”NH₃” = P_”HI” = (nRT)/V = (“0.017 mol” × “0.0821 L·atm·K”^-1″mol”^-1 × “676 K”)/”2.00 L”## = 0.47 atm

4. Calculate ##K_P##.

##K_P = P_”NH₃”P_”HI”## = 0.47 × 0.47 = 0.2

Note: The answer can have only 1 significant figure, because that is all you gave for the starting mass of NH₄I. If you need more precision, you will have to recalculate.