Question

co₂(g) + 2so₂(g) → 3o₂(g) + cs₂(g)
considering the thermodynamic data given below, which statement describes the thermodynamic favorability of the system?
δg° = +1062 kj a. always favorable
δh° = +1005 kj b. never favorable
δs° = +143 j/k c. favorable at high t
d. favorable at low t
enter the answer choice letter.

Explanation:

Step1: Recall Gibbs Free Energy Formula

The spontaneity of a reaction is determined by the Gibbs free energy change, \(\Delta G\), using the formula \(\Delta G = \Delta H - T\Delta S\). A reaction is spontaneous (favorable) when \(\Delta G < 0\).

Step2: Analyze Sign of \(\Delta H\) and \(\Delta S\)

Here, \(\Delta H^\circ = +1005\space kJ\) (endothermic, positive) and \(\Delta S^\circ = +143\space J/K = +0.143\space kJ/K\) (increase in entropy, positive).

Step3: Determine Temperature Dependence

For \(\Delta G\) to be negative:
\[ \Delta H - T\Delta S < 0 \\ 1005 - T(0.143) < 0 \\ T(0.143) > 1005 \\ T > \frac{1005}{0.143} \approx 7028\space K \]
Since \(\Delta H\) is positive and \(\Delta S\) is positive, the reaction becomes favorable ( \(\Delta G < 0\)) at high temperatures (when \(T\) is large enough to make \(T\Delta S > \Delta H\)). But wait, the given \(\Delta G^\circ\) is positive, but we analyze the general case. Wait, no—wait, the standard \(\Delta G^\circ\) is positive, but the formula \(\Delta G = \Delta G^\circ + RT\ln Q\), but for the thermodynamic favorability (spontaneity), we use \(\Delta G = \Delta H - T\Delta S\). Wait, actually, the standard \(\Delta G^\circ\) is at 298 K. Let's check at 298 K: \(\Delta G = 1005 - 298\times0.143 \approx 1005 - 42.6 = 962.4\space kJ > 0\). But as temperature increases, \(T\Delta S\) increases. So when \(T\) is very high, \(T\Delta S\) will exceed \(\Delta H\), making \(\Delta G < 0\). But wait, the options: but wait, the given \(\Delta G^\circ\) is positive, but maybe the question is about the general case. Wait, no—wait, the options are about "thermodynamic favorability" (spontaneity). Wait, but the \(\Delta G^\circ\) is positive, but let's re-express. Wait, maybe I made a mistake. Wait, the formula is \(\Delta G = \Delta H - T\Delta S\). So:

- If \(\Delta H > 0\) and \(\Delta S > 0\): reaction is favorable at high \(T\) (because \(T\Delta S\) can overcome \(\Delta H\)).
- If \(\Delta H < 0\) and \(\Delta S < 0\): favorable at low \(T\).
- If \(\Delta H > 0\) and \(\Delta S < 0\): never favorable (since both terms make \(\Delta G\) more positive).
- If \(\Delta H < 0\) and \(\Delta S > 0\): always favorable.

In this case, \(\Delta H > 0\), \(\Delta S > 0\): so favorable at high \(T\). But wait, the given \(\Delta G^\circ\) is positive, but the question is about "thermodynamic favorability" (spontaneity). Wait, but the options: option C is "Favorable at high T", D is "Favorable at low T". Wait, but let's check the numbers. Wait, \(\Delta H = +1005\space kJ\), \(\Delta S = +0.143\space kJ/K\). So the temperature where \(\Delta G = 0\) is \(T = \Delta H / \Delta S = 1005 / 0.143 \approx 7028\space K\). So above this temperature, \(\Delta G < 0\) (favorable). But the options: but wait, the given \(\Delta G^\circ\) is positive, but maybe the question is using the standard values? Wait, no—maybe I messed up. Wait, the standard \(\Delta G^\circ\) is positive, but the formula for spontaneity is \(\Delta G = \Delta H - T\Delta S\). So with \(\Delta H\) positive and \(\Delta S\) positive, the reaction is spontaneous (favorable) at high temperatures. But wait, the options: option C is "Favorable at high T", but wait, the \(\Delta G^\circ\) is positive, but maybe the question is about the general case. Wait, no—wait, maybe the question has a typo, but according to the signs: \(\Delta H\) positive, \(\Delta S\) positive: so favorable at high T. But wait, the \(\Delta G^\circ\) is positive, but that's at 298 K. So the correct answer should be C? Wait, no—wait, wait, the \(\Delta G^\circ\) is +1062 kJ, which is positive, but the formula \(\Delta G = \Delta H - T\Delta S\). Wait, maybe I confused \(\Delta G^\circ\) with \(\Delta G\). Let's recast:

The spontaneity is determined by \(\Delta G\), not \(\Delta G^\circ\), unless \(Q = 1\) (standard state). But the question is about "thermodynamic favorability" (spontaneity) of the system. So using \(\Delta G = \Delta H - T\Delta S\):

- \(\Delta H\) is positive (endothermic), \(\Delta S\) is positive (increase in disorder). So for \(\Delta G < 0\), we need \(T\Delta S > \Delta H\), which happens at high temperatures (since \(T\) is in Kelvin, positive). So the reaction is favorable at high temperatures. But wait, the \(\Delta G^\circ\) is positive, but that's at 298 K. So the correct option is C? Wait, no—wait, the \(\Delta G^\circ\) is given as +1062 kJ, which is positive, but maybe the question is about the general case. Wait, maybe I made a mistake. Wait, let's check the options again.

Wait, the options:

A. Always favorable: \(\Delta G\) would need to be always < 0. For that, \(\Delta H < 0\) and \(\Delta S > 0\) (since then \(\Delta H - T\Delta S\) would be < 0 for all T). But here \(\Delta H\) is positive, so A is wrong.

B. Never favorable: \(\Delta G\) always > 0. That would be if \(\Delta H > 0\) and \(\Delta S < 0\) (since then \(\Delta H - T\Delta S\) would be > 0 + T|ΔS|, always positive). But here \(\Delta S\) is positive, so B is wrong.

C. Favorable at high T: \(\Delta H\) positive, \(\Delta S\) positive: then at high T, \(T\Delta S\) becomes larger than \(\Delta H\), so \(\Delta G = \Delta H - T\Delta S < 0\). So this is correct.

D. Favorable at low T: \(\Delta H\) positive, \(\Delta S\) positive: at low T, \(T\Delta S\) is small, so \(\Delta H - T\Delta S\) is positive (since \(\Delta H\) is large positive), so \(\Delta G > 0\), not favorable. So D is wrong.

Wait, but the \(\Delta G^\circ\) is positive, but that's at 298 K. The key is the signs of \(\Delta H\) and \(\Delta S\). So with \(\Delta H\) positive and \(\Delta S\) positive, the reaction is favorable at high temperatures. So the correct answer is C? Wait, but wait, the \(\Delta G^\circ\) is +1062 kJ, which is positive, but the formula for \(\Delta G\) is \(\Delta H - T\Delta S\). So when T is very high, \(\Delta G\) becomes negative. So the answer should be C. But wait, the options: C is "Favorable at high T". So that's the correct option.

Answer:

C. Favorable at high T