The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are, –890.3 kJ mol –1 –393.5 kJ mol –1 , and –285.8 kJ mol –1 respectively. Enthalpy of formation of CH 4 (g) will be (i) –74.8 kJ mol –1 (ii) –52.27 kJ mol –1 (iii) +74.8 kJ mol –1 (iv) +52.26 kJ mol –1

Question

Philoid · Accepted Answer

Given,

We need to find enthalpy change for reaction

ΔH₅ = ΔH(CH₄)

ΔH₁ = ΔH(CO₂) + 2ΔH(H₂O) – ΔH(CH₄)

ΔH₂ = ΔH(CO₂)

ΔH₄ = 2ΔH(H₂O)

Thus,

ΔH(CH₄) = ΔH₂ + ΔH₄ – ΔH₁

= -393.5 + (-571.68) + 890.3

= - 73.8 KJ

Thus, answer is (i) –74.8 kJ mol^–1

The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are, –890.3 kJ mol–1 –393.5 kJ mol–1, and –285.8 kJ mol–1 respectively. Enthalpy of formation of CH4(g) will be(i) –74.8 kJ mol–1 (ii) –52.27 kJ mol–1(iii) +74.8 kJ mol–1 (iv) +52.26 kJ mol–1

The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are, –890.3 kJ mol^–1 –393.5 kJ mol^–1, and –285.8 kJ mol^–1 respectively. Enthalpy of formation of CH₄(g) will be
(i) –74.8 kJ mol^–1 (ii) –52.27 kJ mol^–1

(iii) +74.8 kJ mol^–1 (iv) +52.26 kJ mol^–1