Consider the reaction: C2H4(g) + H2(g) <-> C2H6(g). For this process, deltaH = -
ID: 948670 • Letter: C
Question
Consider the reaction: C2H4(g) + H2(g) <-> C2H6(g). For this process, deltaH = -137.0 kj/mol and deltaS = -120.6 J/mol*K. Based on these data and assuming deltaH and deltaS are temperature independent, answer the following questions;
1) Does this reaction favor products or reactants at 25 degrees C?
2) At 25 degrees C, is the reaction driven(dominated) by the entropy or the enthalpy?
3) Would a decrease in temperaturefavor the reactants or the products?
4) At what temperature would the equilibrium constant (K) equal unity (=1.0)?
The answers should be 1:products, 2:enthalpy, 3:products and 4:???? but I don't know why! Explaination please and answer to #4.
Explanation / Answer
C2H4(g) + H2(g)----------> C2H6(g)
Has D(H) = -137 kJ/mol = -137000 J/mol (D means Delta)
D(S) = -120.6 J/mol. K and T = 25oC = 298 K
We know that the D(G), D(H), D(S0 and T related as,
D(G) = D(H) -T D(S).
D(G) = -137000 - 298 x (-120.6)
D(G) = -101061.2 J/mol
D(G) = -101.061 kJ/mol
Sign of D(G) is -ve and hence the reaction is spontaneous and will proceds in forwad direction i.e Product formation os favored.
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2. This reaction at 25oC will be enthalpy driven
As we can see that both D(H) and D(S) are negative and hence when both D(S) and D(H) are -ve the reaction is said to be Enthalpy driven,
(NOTE: when both have +ve sign then reaction is entropy driven. And when signs of D(S) and D(H) are different they neither of them drive the reaction.)
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3) If we look at the calculation done in answer of Q1 its observed that as we lower the Temperature T the product T x D(S) become small (remains +ve only) and hence its addition to -137000 gives more -ve value of D(G) and reaction will become more spontaneous in forward direction and Product formation will be favored.
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4) D(G), R, T and ln(K) where K is equilibrium constant are related as,
D(G) = -RT ln(K)
We want to know that T = ? for ln(K) = 1 {not K may be) and D(G) = -101061.2 J/mol and R (molar gas constant = 8.314 J/mol.K
Let us put these values in the above equation and solve it for T
-101061.2 = -8.314 x T x 1
T = -101061.2/ -8.314
= 12155 K (apprx.)
At such high temperature ln(K) will be unity.
Or If we still want K = 1
We get ln(K) = ln(K) = o means, D(G) / T = 0 (8.314 ignored)
This is possible only if T = infinity as we know that D(G) is not 0 .
So theoretically speaking at infinte temperature equilibrium constant is 1.
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