Question

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

Answer 1

Here, the amount of NH₄Cl is not that useful because it is in solid state.

Thus,

$K=[NH_3][HI]$

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	$\rightarrow$	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,

$7.00\times 10^{-5}=(1.46\times10^{-2}-x)(8.37\times 10^{-3} -x)\\ \Rightarrow 0.00007 = x^2 - 0.02297 x + 0.000122202\\ \Rightarrow -x^2 + 0.02297 x - 0.000052202 = 0\\ \Rightarrow x=0.00255733\;or\;0.0204127$

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