# Change in Entropy and Enthalpy found over a temperature range and asked to compute the change in Gibb's Energy

I was given the following question by my tutor:

"The variation of enthalpy ($$\Delta H$$) and entropy ($$\Delta S$$) in a reaction carried out at constant pressure in the temperature range between 300 ºC and 350 ºC, are 1700 kcal and 15kcal/K respectively. Calculate the range of temperatures in which this reaction is spontaneous."

But my book on thermodynamics says that the change in Gibb's energy for a process can only be calculated if we start and end at the same temperature (the temperature can vary during the process as heat is added or lost from the system to the surroundings, but the end points must have constant temperature and pressure.) In this question the changes in entropy and enthalpy are found over a range of temperatures where the final temperature is not the same as the initial temperature, but the pressure is always constant.

Is my tutor's question invalid, is my book wrong in saying that the temperature has to be the same at the initial and final values, or am I thinking about this wrong?

• $\ce{\Delta G}$ can be calculated by the expression ${\Delta G = \Delta H - T \Delta S}$. Why don't you calculate the change of H and S, then put ${\Delta G = 0}$. This an equation with one unknown T. A simple substraction plus a division gives you the minimum temperature for the reaction to become spontaneous. – Maurice Mar 30 '20 at 17:04
• @Maurice Yes, this is the "solution" I was given, but my question is not regarding answering the question, it's whether the question is valid since the enthalpy and entropy were taken over a range of temperatures rather than at constant end points of temperature and pressure (Gibb's function is defined so that a spontaneous process with starting pressure $P_0$ and temperature $T_0$ and the same pressure $P_0$ and temperature $T_0$ at the end point yields a $\Delta G < 0$, according to my book!) – SON1 Mar 30 '20 at 17:16
• @ Son1. You are right in stating that in order to calculate $\Delta G$, the temperature of the system must be the same in the beginning and at the end of the reaction. But it does not mean that this temperature must stay at a given value. The reaction may occur at any temperature, provided it is the same at the beginning and at the end of the reaction. $\Delta G$ can be calculated at any temperature, provided you know the change in enthalpy and entropy, and provided these changes in H and S are not changing with the temperature. – Maurice Mar 30 '20 at 19:49
• I think what is (too loosely) being referred to in the problem statement is the standard changes in enthalpy and entropy for the reaction. I think the intent is for you to assume that they are nearly constant over the temperature range from 300 C to 350 C. And I think you are supposed to use the rule of thumb that a reaction is spontaneous at a given temperature if the change in Gibbs free energy is less than zero. – Chet Miller Mar 30 '20 at 20:27

$$G = H - TS \Rightarrow dG = dH -TdS - SdT$$

$$\Rightarrow \text{(at constant T) } dG = dH -TdS \Rightarrow \Delta G = \Delta H - T \Delta S$$

So, yes, you need $$\Delta T = 0$$ for $$\Delta G = \Delta H - T \Delta S$$ to hold.

However, this holds at any T.

E.g., suppose you do the reaction at $$300 K$$. $$\Delta T = 0$$, so $$\Delta G = \Delta H - T \Delta S$$ holds.

But suppose you repeat reaction at $$350 K$$. It's still the case that $$\Delta T = 0$$, so $$\Delta G = \Delta H - T \Delta S$$ still holds.

As far as the question itself goes, I agree with Chet Miller that it appears to be badly worded. It seems what your tutor meant to say was:

"Between 300 ºC and 350 ºC, assume that $$\Delta H$$ and $$\Delta S$$ are constant, with values of 1700 kcal and 15kcal/K, respectively. Calculate the range of temperatures over which $$\Delta G <=0$$, i.e., the range the textbook labels as 'spontaneous'".