At 25°C,?G°_fis -620 kJ/mol for SiCl₄(g)and -592 kJ/mol for MgCl₂(s).Calculate ?G° for the reaction,SiCl₄(g)+ 2 MG(s) $\rarr$ 2 MgCl₂(s)+ Si(s)and determine if the reaction is spontaneous at 25°C if the pressure of SiCl₄(g)is 1 atm.

Question

Quizplus · Accepted Answer

The equation for the reaction is:
SiCl4(g) + 2 Mg(s) → 2 MgCl2(s) + Si(s)

The overall ΔG° for the reaction can be calculated using the formula:
ΔG° = ∑ΔG°f(products) - ∑ΔG°f(reactants)

From the given data, we have:
ΔG°f(SiCl4(g)) = -620 kJ/mol
ΔG°f(MgCl2(s)) = -592 kJ/mol
ΔG°f(Si(s)) = 0 kJ/mol
ΔG°f(Mg(s)) = 0 kJ/mol

Using the above values, we can calculate the ΔG° for the reaction as follows:

ΔG° = [2 × ΔG°f(MgCl2(s))] + [ΔG°f(Si(s))] - [ΔG°f(SiCl4(g))] - [2 × ΔG°f(Mg(s))]

ΔG° = [2 × (-592 kJ/mol)] + [0 kJ/mol] - [-620 kJ/mol] - [2 × 0 kJ/mol]

ΔG° = -1184 kJ/mol + 620 kJ/mol = -564 kJ/mol

Since ΔG° is negative, the reaction is spontaneous at 25°C. Therefore, the correct answer is C.

At 25°C,?G°f is -620 KJ/mol for SiCl4(g)and -592 KJ/mol for MgCl2(s).Calculate

At 25°C,?G°_fis -620 KJ/mol for SiCl₄(g)and -592 KJ/mol for MgCl₂(s).Calculate