The equilibrium constant, K, for the following reaction is 2.57×10-4 at 549 K.
NH4Cl(s) NH3(g) + HCl(g)
An equilibrium mixture in a 14.3 L container at 549 K contains 0.321 mol NH4Cl(s), 1.97×10-2 M NH3 and 1.30×10-2 M HCl.
What will be the concentrations of the two gases once equilibrium has been reestablished, if the equilibrium mixture is compressed at constant temperature to a volume of 6.96 L?
[NH3] = __________M
[HCl] = ___________M
Answer:
NH4Cl(sNH3(g) + HCl(g)
Initially,molar grouping of NH3 = moles of NH3/Volume of soltion in liters = 0.0197/6.96 = 0.00273 M
Initially,molar grouping of HCl = moles of HCl/Volume of soltion in liters = 0.0130/6.96 = 0.00187 M
Therefore, Reaction Quotient,Kq = [NH3]*[HCl] = 5.1051*10-6
Since, Kq < Keqb , therefore, more items will be shaped
Let at eqb., [NH3] = 0.00273+x and [HCl] = 0.00187+x
In this way, Keq = [NH3]*[HCl] = (0.00273+x)*(0.00187+x)
or on the other hand, 2.57*10^-4 = x2 + 0.0046x+0.0000051051
Subsequently, x2 + 0.0046x - 0.000252 = 0
or then again, x = 0.01374
Subsequently, at eqb., [NH3] = 0.01647 M
[HCl] = 0.01561 M
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