Question

Heat of combustion ΔH for C(s) , H₂(g)and CH₄(g) are -94, -68 and -213 kcal/mol. Then ΔH for C(s) + 2H₂(g) → CH₄(g)is

Answer 1

QUESTION

• Heat of combustion ΔH for C(s) , H₂(g)and CH₄(g) are -94, -68 and -213 kcal/mol. Then ΔH for C(s) + 2H₂(g) → CH₄(g) is

EXPLANATION

GiVEN

For C(s) + O₂ = CO₂

ΔH = -94 kcal/mol

For 2H₂ + O₂ = 2H₂O

ΔH = -68 x 2 kcal/mol

For CH₄ + 2O₂ = CO₂ + 2H₂O

ΔH = -213 kcal/mol

To FiND

∆H for C(s) + 2H₂(g) = CH₄(g)

SOLUTiON

Assume

C(s) + O₂ = CO₂ _(i)

2H₂ + O₂ = 2H₂O _(ii)

CH₄ + 2O₂ = CO₂ + 2H₂O _(iii)

C(s) + 2H₂(g) = CH₄(g) _(iv)

Applying Hess’s law

ΔH(iv), (i) + (ii) – (iii), we have

C + O₂ + 2H₂ + O₂ – CH₄ – 2O₂ = CO₂ + 2H₂ O – CO₂ – 2H₂ O

C + 2H₂ = CH₄

Therefore

ΔH(iv) = ΔH(i) + ΔH(ii) – ΔH(iii)

= (-94 – 68 x 2 + 213 ) kcal

= -17 kcal.

Answer 2

Answer:

C(s)+2H2(g)→CH4(g),ΔH°=?

ΔH°= −[(ΔH° of combustion of CH4) - (ΔH° of combustion of  C)+ (2 ×ΔH° of combustion of H2)]

C+O2→CO2;ΔH°=−94kcal ... (i)

2H2+O2→2H2O;ΔH°=−68×2kcal ... (ii)

                   =−[(−213)−(−94+2×−68)]kcal/mol

                   =−[−213+230]=−17kcal/mol