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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: slyasafax on April 08, 2022, 08:18:47 PM

I've been stuck with this problem:
The following reaction has Kc = 2.8 · 10^{2} at T=1000 K:
C_{2}H_{4} + H_{2} ::equil:: C_{2}H_{6}
A reactor with volume V = 10 L initially contains 0.32 mol C_{2}H_{4}, 0.16 mol H_{2} and 0.68 mol C_{2}H_{6}.
Calculate the number of C_{2}H_{6} moles after:
a) the addition of 1.00 mol of C_{2}H_{6} in the reactor.
b) the volume of the reactor is increased by 20 L.
I know the answer to a) must be 1.515 moles, while b) must be 0.645 moles.
I can see these numbers make sense by LeChatelier principle. By adding 1 mole to 0.68 mol you get 1.68 mol of C_{2}H_{6}, but a part of it gets converted back to its reactants; by adding 20 L to the volume of the recipient, you shift the equilibrium position towards the reactants (because they are gaseous, so they are entropically favoured)
I tried to solve this exercise with ICE tables, resulting in
Numerator: 1.68 + x
Denominator: (0.32  x) × (0.16 x)
But when I solve for Numerator/Denominator = Kc, I do not get the answers above. I also do not know how to include volume in my calculation. I've tried using the ideal gas law and converting into partial pressures, but I still do not get the correct answer.

Might the first result (point a) contain an error?
The way I've gone about this problem is this: define Kp in terms of Kc. Use the fact that, for a reaction with two reagents and one product, all of them with stoichiometric coefficient = 1:
Kp = Kc/RT
So you get
Kc/RT = [n_{C}RT/V] / [n_{A}·n_{B}·(RT/V)^{2}]
by eliminating all of the RT's, this massively simplifies to
Kc/V = n_{C} / n_{A}·n_{B}
at this point you can calculate what happens to the ratio if you increase V by 20 L. Simply substitute n_{C} for n_{C}x, n_{A}·n_{B} for (n_{A}+x)·(n_{B}+x), and 10 for 30 in the denominator of Kc. By doing this you get n_{C}x ≈ 0.645.
In point a), it is de facto impossible for the moles of H2 to decrease by 0.165 (1.68  1.515 of the alleged result) because they are 0.16 to begin with.

In point a), it is de facto impossible for the moles of H2 to decrease by 0.165 (1.68  1.515 of the alleged result) because they are 0.16 to begin with.
Moles of H_{2} don't decrease, they increase. As you said, Le Chatelier.