Question

what is the minimum temperature for the following reaction to occur?

CaO + SiO2 --> CaSiO3

Thanks in advance.

Answer 1

Answer on Question #60855 - Chemistry - Other

Task:

What is the minimum temperature for the following reaction to occur?

CaO + SiO2 --> CaSiO3

Solution:

$$CaO + SiO_2 \rightarrow CaSiO_3.$$

The standard-state enthalpy of reaction is equal to the sum of the enthalpies of formation of the products minus the sum of the enthalpies of formation of the reactants:

$$\Delta H^o(\text{reaction}) = \sum \Delta H_f^o(\text{products}) - \sum \Delta H_f^o(\text{reactants}).$$ $$\Delta H^o(\text{reaction}) = \Delta H_f^o(CaSiO_3) + \Delta H_f^o(SiO_2) - \Delta H_f^o(CaO).$$ $$\Delta H^o(\text{reaction}) = -1584.1 - (-859.3) - (-635.1) = -89.7(kJ \times mol^{-1}).$$

The standard entropy of reaction is equal to the sum of the entropies of the products minus the sum of the entropies of the reactants:

$$\Delta S(\text{reaction}) = \sum \Delta S(\text{products}) - \sum \Delta S(\text{reactants}).$$ $$\Delta S(\text{reaction}) = 82 - 42.1 - 39.7 = 0.2(J \times mol^{-1}).$$

For a favorable reaction ($\Delta G < 0$), $|\Delta H|$ must be smaller than $|T\Delta S|$, or $\Delta G$ must go through zero.

So the temperature where $\Delta G$ is zero is the temperature where the spontaneity changes.

$$\Delta G = \Delta H - T\Delta S$$ $$0 = \Delta H - T\Delta S$$ $$\Delta H / \Delta S = T$$ $$T = 89700 \text{ J/mol} / (0.2 \text{ J/mol K}) = -448500 \text{ K}$$

So, T must be greater than -448500 K