Question

Consider the reaction I2(g) + Cl2(g)2ICl(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 2.02 moles of I2(g) react at standard conditions. S°surroundings = J/K

Answer 1

Given:

Hof(I2(g)) = 62.438 KJ/mol

Hof(Cl2(g)) = 0.0 KJ/mol

Hof(ICl(g)) = 17.78 KJ/mol

Balanced chemical equation is:

I2(g) + Cl2(g) ---> 2 ICl(g)

ΔHo rxn = 2*Hof(ICl(g)) - 1*Hof( I2(g)) - 1*Hof(Cl2(g))

ΔHo rxn = 2*(17.78) - 1*(62.438) - 1*(0.0)

ΔHo rxn = -26.878 KJ/mol

This is when 1 mol of I2 reacts.

For 2.02 mol of I2,

ΔHo rxn = -26.878 KJ/mol * 2.02 mol = - 54.29 KJ = -54290 J

Use:

So surr = -ΔHo rxn / T

= -(-54290 J) / 298 K

= 182 J/K

Answer: 182 J/K