Equilibrium at a temperature of 82 degreeC inside a sealed fixed-volume chamber,
ID: 957491 • Letter: E
Question
Equilibrium at a temperature of 82 degreeC inside a sealed fixed-volume chamber, there was found to be 0.080 atm of gas A present, 0.270 atm of gas B present and 0.380 atm of gas D present. If more of gas D is pumped into the chamber at constant temperature, such that the new pressure of gas D is 0.100 atm higher than it was before, what is the deltaG of the reaction below now? Assume the extra D gas was pumped in so quickly that during the pumping no amount of reaction occurred during the pumping process. Which way will the reaction shift to regain equilibrium after the pumping process is complete? A(g) + 3B(g) right arrow 2D(g)Explanation / Answer
Answer – We are given, At equilibrium , P of A = 0.080 atm , P of B = 0.270 atm and P of D = 0.380 atm
Reaction – A (g) + 3B(g) <----> 2D(g)
Calculation of Kp –
We know, Kp = P(D)2 / P(A)*P(B)3
= (0.380)2 / (0.080)(0.270)3
= 91.7
Now when we increase the pressure of D by 0.100 atm than original then reaction gets reversed and there is got a new Kp
At equilibrium, P of D = 0.380 + 0.100 = 0.480 atm
2D(g) <----> A (g) + 3B(g)
Qp = P(D)2 / P(A)*P(B)3
= (0.480)2 / (0.080)(0.270)3
= 146.3
We know,
Go = -RT*ln K
= - 8.314 J/mol.K * 355 K * ln 91.7
= -13336.3 J
We know,
G = Go + RTln Qp
= -13336.3 J + 8.314 J/mol.K * 355 K * ln 146.3
= 1379.1 J
= 1.38 kJ/mol
So, G is positive and then reaction will shift towards right.
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