Question

13.(6 pts) The In Ke, for the reaction below was deternined through a range of temperatures and inie plotted inverse temperature. The graph is shown, with the equation of the function. SO2 (g) O2(g) SO3 (g) + → (a) Determine AHo and AS for this process In Keq as a function of Inverse Temperature 2.10E+02 1.90E+02 1.70E-02 1.50E+02 1.30E+02 1.10E-02 E 9.00E+01 Z 7.00E+01 5.00E+01 3.00E+01 2381 y 23814x - 22.491 R2 1 0.008 0.01 0.002 0.004 0.006 Inverse temperature, 1/Kelvin (b) In what temperature range is this reaction spontaneous (product favored)? Circle one low temps only Il tempsno temphigh temps only transition occurs between spontaneous and non-spontaneous.

Answer 1

From the above calculations

C) τ @ дн°-1AS --RTL, Kei 21 - RT △RF R1 R. AS"--足2.49, △g": -22-491R slo -A31 R. R. K1尺

(a) delta S = -22.491R and deltaH= -23814R.

(b) for the reaction to be spontaneous, delta G should be negative, and delta G is negative at high temperatures above 1108K.

(c)The temperature at which the transition from non-spontaneous to spontaneous reaction takes place is at 1108K.