The following graph depicts the realtionship ?G° vs T on the basis of data colle
ID: 1016370 • Letter: T
Question
The following graph depicts the realtionship ?G° vs T on the basis of data collected experimentally for some reaction:
a) determine the slope of the line that fits the data in the graph. (---Pick two points far apart from each other)
b) calculate ?S° for this reaction (the sign matters). (My professor had said ?S° would be negative ...)
c) Calculate ?H° for this reaction (the sign matters).
d) is this reaction endothermic or exothermic?
e) does this reaction become more favorable at higher T or lower T?
PLEASE EXPLAIN, Thank you!
Explanation / Answer
We know that G0 = H0 – TS0 where H0 is the enthalpy change and S0 is the entropy change
(a) Since I do not have the exact scale, I will presume values. The lowest point is (y = 1.8 kJ, x = 290 K) and let us take another point near the top (y = 7.5 kJ, x = 330 K; I took the highest but one point).
Slope of the curve is m = (y2 – y1)/(x2 – x1) = (7.5 – 1.8)/(330 – 290) = 0.1425 (ans)
(b) For this part, compare the parent equation G0 = -TS0 + H0 with the equation of a straight line y = mx + c. T is positive. Therefore, we must have m = -S0 and c = H0. Therefore, S0 = -m = -0.1425 kJ/K = -142.5 J/K (the sign is negative as your professor said) (ans)
(c) The y-intercept will give the value of c = H0. However, we do not see a y-intercept on the positive y-axis which means that there is an intercept on the negative y-axis. That is not shown in the figure, so we will simply use our parent relation G0 = H0 – TS0 at a particular temperature (say 330 K) and calculate H0. We must assume however that H0 and S0 are independent of temperature and hence constant. Therefore, from the graph,
7.5 kJ = (330 K)*(-142.5 J/K) + H0 = -47.025 kJ + H0
===> H0 = 7.5 kJ + 47.025 kJ = 54.525 kJ (ans)
(d) The sign of H0 is positive while S0 is negative. Overall, G0 is positive and the reaction is endothermic.