Question

I dont know how to explain that SnO formation from its elements is thermodynamically favorable. I think it must be redox because thats how we explain thermodynamic favorable equations. But i dont know if i can use water molecules because is says From its elements. thank you :)

Answer 1

I don't know, how to explain that SnO formation from its elements is thermodynamically favorable. I think it must be redox because that's how we explain thermodynamic favorable equations. But I don't know if I can use water molecules because it says "from its elements".

Solution: Such explanation should be based on the sign of the Gibbs free energy change for the reaction of SnO formation from elements in standard state (298 K, 1.013 bar):

$$2 \mathrm {S n} _ {\text {(s o l i d)}} + \mathrm {O} _ {2} (\mathrm {g a s}) \rightarrow 2 \mathrm {S n O} _ {\text {(s o l i d)}}$$

$\Delta G = \Delta H_{r} - T \cdot \Delta S_{r}$ , where $\Delta G$ - Gibbs free energy change, kJ, if $\Delta G < 0$ , then this reaction is thermodynamically favorable; $\Delta H_{r}$ - reaction enthalpy change, kJ, $\Delta H_{r} = \Sigma (\Delta_{f}H_{i}^{0}\cdot n_{i})_{products} - \Sigma (\Delta_{f}H_{i}^{0}\cdot n_{i})_{reagents}$ , where $\Delta_{f}H^{0}$ - standard enthalpy change of formation of substance, kJ/mol; n - substance amount, mol; T - absolute temperature, K; $\Delta S_{r}$ - reaction entropy change, kJ/K, $\Delta S_{r} = \Sigma (S_{i}^{0}\cdot n_{i})_{products} - \Sigma (S_{i}^{0}\cdot n_{i})_{reagents}$ , where $S^{0}$ - standard entropy of formation of substance, kJ·mol $^{-1}$ ·K $^{-1}$ ; n - substance amount, mol; All the values of $\Delta_{f}H^{0}$ and $S^{0}$ are available in handbooks.

We will not calculate the standard Gibbs free energy change of SnO formation, because it is too available in the handbooks, and it has value of $-257.3\mathrm{kJ / mol}$ , as you can see, $\Delta G < 0$ , the reaction of SnO formation from its elements is thermodynamically favorable.

Answer: The standard Gibbs free energy change of SnO formation is negative, and then the reaction of SnO formation from its elements is thermodynamically favorable.