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I saw this reaction in my textbook:

$\ce{NH4Cl(s) -> NH3(g) + HCl(g)}$ given $\Delta{H}_{r} = \pu{176 kJ/mol}$ and $\Delta{S} = \pu{0.285 kJ/(mol \cdot K)}$

On a worksheet my teacher gave me, I saw the reverse reaction:

$\ce{NH3(g) + HCl(g) -> NH4Cl(s)}$ given $\Delta{H}_{r} = \pu{-176 kJ/mol}$ and $\Delta{S} = \pu{-0.285 kJ/(mol \cdot K)}$

Both reactions are said to occur at $\pu{298.15 K}$ and $\pu{1 atm}$ of pressure.

I know that when reversing the reaction, the sign of $\Delta{H}_{r}$ must be reversed as well. Doing the same for $\Delta{S}$ makes sense as well due to the state change.

The change in free energy in the first reaction is $\pu{+91 kJ/mol}$, whereas in the second it is $\pu{-91 kJ/mol}$.

Would it be correct to infer from this, that when an irreversible reaction is spontaneous, its reverse is nonspontaneous (under the same conditions) and vice versa?

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  • $\begingroup$ Direction of spontaneity and the value of reaction Gibbs energy depends on partial pressures of gases. $\endgroup$
    – Poutnik
    Commented Dec 22, 2023 at 14:42
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    $\begingroup$ Yes, it is correct to infer this, but only if the substances are in their standard state for which the values of $\Delta H$ and $\Delta S$ are given. For other temperatures and partial pressures, as mentioned by @Poutnik, you have to use $$\Delta G = \Delta G^\circ + RT \ln Q$$ where $Q$ is the reaction quotient. $\endgroup$
    – Domen
    Commented Dec 22, 2023 at 14:52

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