The equilibrium constant, K, for the following reaction is
7.00×10-5 at 673 K.
NH4I(s)----> NH3(g) + HI(g)
An equilibrium mixture of the solid and the two gases in a 1.00 L flask at 673 K contains 0.388 mol NH4I, 8.37×10-3 M NH3 and 8.37×10-3 M HI. If the concentration of NH3(g) is suddenly increased to 1.46×10-2 M, what will be the concentrations of the two gases once equilibrium has been reestablished?
[NH3] = ______ M
[HI] = _______ M
Here, the amount of NH4Cl is not that useful because it is in solid state.
Thus,
From the reaction, it is evident that NH3 and HI reacts in the ratio 1:1.
Let x moles of NH3 reacts. We will use ICE table
Reaction | NH4Cl | NH3 | HI | |
Initial | Not needed | 1.46 x 10-2 | 8.37 x 10-3 | |
Change | Not Needed | -x | -x | |
Equilibrium | Not Needed | 1.46 x 10-2 - x | 8.37 x 10-3 - x |
Thus, at equilibrium, the products of concentration will equal K. Hence,
But if we take x= 0.0204127, then [HI] = 0.00837 - 0.0204127 M = -0.0120427. But [HI] cannot be negative.
Hence, x = 0.00255733 = 2.56 x 10-3 (approx)
[NH3]= 1.46 x 10-2 - 2.56 x 10-3 M= 1.20 x 10-2 M
[HI]= 8.37 x 10-3- 2.56 x 10-3 M= 5.81 x 10-3 M
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