Question

100 200 300 400 500 600 Temperature (K) 1. Describe the relationship between the spontaneity of the process and temperature. 2. Is the reaction exothermic? 3. Is AS >O? 4. At what temperature is the reaction at standard conditions likely to be at equilib- rium? 5. What is K for the reaction at 27°C? C. At what temperature will acetone boil if AHvap = 30.3 kJ/mol, and A Svap = 92.0 J/mol- K? D. Consider the reaction

need help with question b and c. please show work

Answer 1

1) Spontaneity of the reaction increses with increase in temperature. Since  ΔG is a measure of spontaneity of a reaction ,in graph the ΔG value become more and more negative as the tmperature increases. A more negative free energy value shows increased spontaneity.

2) Yes the rection may be exothermic.

We have the equation

AG = AH-TAS

For a reaction to be spontaneous ΔG should be negative and there are different conditions

(a) ΔH is negative (exothermic),ΔS is positive, At this condtion ΔG will be always negative at all temperature

(b) ΔH is negative ,ΔS is negative, this become spontaneous when ΔH>TΔS

(c) ΔH is positive (endothermic) ,  ΔS negative, always non-spontaneous

(d) ΔH is positive, ΔS is positive, spontaneous when ΔH<TΔS

Here ΔG become more negative with increasing temperature. In the case of exothermic reaction ΔH is negative and if ΔS is positive. Both the term in equation is negative and ΔG become negative. The negative value increases with temperature.In the case of exothermic reaction whose entropy increases will be spontaneous at all temperatures. But here from graph at lower temperature the value of ΔG is positive. It may be due to decresed entropy at lower temperature and the value of -TΔS (which is positive) exceeds ΔH (negative for exthermic reaction ) and ΔG become positive. As the temperature increases, randomness increases and ΔS become positive.

3)As the temperature increases randomness increases and ΔS0>0.

4) At equilibrium ΔG0 = 0

ΔH0=TΔS0

Here from the graph, the temperature is 320 K

(4) ΔG0 = -RTlnK

At 270C (300K), ΔG0=10 KJ/mol = 10$\times$$10^{3}$ J/mol R=8.314 JK-1mol-1

lnK = -10$\times$$10^{3}$/8.314$\times$300

=-0.0040$\times$$10^{3}$

K = Antilog (-4)

= 0.0183

(5) Boiling occurs at a temperature were acetone (liquid) is equilibrium with acetone (vapour). This is the equilibrim condition, where ΔG0 = 0

ΔH0=TΔS0

Temperature = ΔHvap/ΔSvap

= 30.3$\times$$10^{3}$J/mol/92 J/K

=329.3K