From the data given below at 298 K for the reaction: CH4(g) + 2O2(g)

1.	From the data given below at 298 K for the reaction: CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) Calculate the enthalpy of formation of CH4(g) at 298 K. Enthalpy of reaction is = -89.5 kJ Enthalpy of formation of CO2(g) = -393kJ mol-1 Enthalpy of formation of H2O(l) = -286.0 kJ mol-1
Answer» ΔH = AH_fCO₂(g) + 2ΔH_fH₂O(l) – ΔH_fCH₄(g) – ΔH_f;O₂(g) -890.5 kJ = -393.5 kJ +2 × -286 kJ - Δ H_fCH₄(g) - 0 ΔH_fCH₄ - 75.0 kJ

From the data given below at 298 K for the reaction: CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) Calculate the enthalpy of formation of CH4(g) at 298 K. Enthalpy of reaction is = -89.5 kJ Enthalpy of formation of CO2(g) = -393kJ mol-1 Enthalpy of formation of H2O(l) = -286.0 kJ mol-1

Answer»

ΔH = AH_fCO₂(g) + 2ΔH_fH₂O(l) – ΔH_fCH₄(g) – ΔH_f;O₂(g)

-890.5 kJ = -393.5 kJ +2 × -286 kJ - Δ H_fCH₄(g) - 0

ΔH_fCH₄ - 75.0 kJ

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From the data given below at 298 K for the reaction: CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) Calculate the enthalpy of formation of CH4(g) at 298 K. Enthalpy of reaction is = -89.5 kJ Enthalpy of formation of CO2(g) = -393kJ mol-1 Enthalpy of formation of H2O(l) = -286.0 kJ mol-1

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