Question

calculate the standard free energy change delta G for reaction N2 (g) +3H2(g)—>2NH3

N2 delta H=0.00kj mol^-1s=+191.5J mol^-1K^-1

H2 delta H=0.00kj mol^-1,s = +130.6j mol^-1 k-1

NH3 delta H=-46.0kj mol^-1,s =192.5 J mol^-1 k-1

A. +112.3 kJ

B.-87.6kJ

C.-7.4kJ

D.-32.9 kJ

E.-151.1kJ

Answer 1

we have:

Hof(N2(g)) = 0.0 KJ/mol

Hof(H2(g)) = 0.0 KJ/mol

Hof(NH3(g)) = -46.0 KJ/mol

we have the Balanced chemical equation as:

N2(g) + 3 H2(g) ---> 2 NH3(g)

deltaHo rxn = 2*Hof(NH3(g)) - 1*Hof( N2(g)) - 3*Hof(H2(g))

deltaHo rxn = 2*(-46.0) - 1*(0.0) - 3*(0.0)

deltaHo rxn = -92 KJ

we have:

Sof(N2(g)) = 191.5 J/mol.K

Sof(H2(g)) = 130.6 J/mol.K

Sof(NH3(g)) = 192.5 J/mol.K

we have the Balanced chemical equation as:

N2(g) + 3 H2(g) ---> 2 NH3(g)

deltaSo rxn = 2*Sof(NH3(g)) - 1*Sof( N2(g)) - 3*Sof(H2(g))

deltaSo rxn = 2*(192.5) - 1*(191.5) - 3*(130.6)

deltaSo rxn = -198.3 J/K

deltaHo = -92.0 KJ/mol

deltaSo = -198.3 J/mol.K

= -0.1983 KJ/mol.K

T = 298 K

we have below equation to be used:

deltaGo = deltaHo - T*deltaSo

deltaGo = -92.0 - 298.0 * -0.1983

deltaGo = -32.9 KJ

Answer: D