At 2200 K the reaction CaO(s) + SO3(g) = CaSO4(s) is nonspontaneous, whereas it is spontaneous at room tempera?

Assuming ΔH and ΔS are both independent of temperature (not a particularly good assumption for this large temperature range), then all the statements in the question are true except for (B).

You should know that ΔG = ΔH - T*ΔS, and the condition for spontaneity is that ΔG < 0.

If the reaction is not spontaneous at 2200K, and spontaneous at 298K (room temperature), then at 2200K, ΔG > 0, and at 298K, ΔG < 0. In order to go from being positive to negative, somewhere in this temperature range, ΔG must equal zero. "A" is therefore true,

The fact that the reaction is spontaneous at room temperature means that ΔG < 0 at room temperature. "C" is true.

The fact that ΔG *increases* as T increases (it goes from being negative to positive at T increases) means that dΔG/dT is positive (i.e., the slope of a plot of ΔG vs. T is positive). Assuming ΔH and ΔS are constant with respect to temperature, we have that:

dΔG/dT = d(ΔH - T*ΔS)/dT = -ΔS > 0

This implies that ΔS < 0, which means that the entropy change acts to *reduce* the spontaneity of the reaction. "B" is therefore false, which means that ΔH is the main driver for the reaction, and "D" is true.

We now know that ΔS < 0, so the entropy term in ΔG = ΔH - TΔS (i.e. -TΔS must be positive, because T is measured in kelvins and is always positive). This means that in order for the reaction to ever be spontaneous, ΔH must be < 0. "E" is therefore true.

The reaction changes from being spontaneous to nonspontaneous (i.e., the equilibrium temperature) at the temperature for which ΔG =0:

ΔG = 0 = ΔH - (T_equil)*ΔS

T_equil = ΔH/ΔS

Cao So3