Standard enthalpy of combustion of $C H_{4}$ is $- 890 k J m o l^{- 1}$ and standard enthalpy of vaporisation of water is 40.5 kJ $m o l^{- 1}$ . Calculate the enthalpy change for the reaction :
$C H_{4 (g)} + 2 O_{2 (g)} \to C O_{2 (g)} + 2 H_{2} O_{(g)}$

Q. The enthalpies of atomization of

C H_{4} (g)

and

C_{2} H_{6} (g)

are respectively

400 k J m o l^{- 1}

and

670 k J m o l^{- 1} .

The value of

△ H_{C - C}

would be

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The heat of formations for C02(g),H20(1) and CH4(G) are −400kJmol−1,−280kJmol−1 and −70kJmol−1 respectively. The heat of combustion of CH4 in mol−1 is?

The correct option is A 890CH4+2O2⟶CO2+2H2O;△CH△CH=∑△fH(reactants)−∑△fH(products)=△fH(CH4)+2△fH(O2)−△fH(CO2)−2△fH(H2O)=−70+(2×0)+400+(2×280)=890 KJ/mol⟹Heat of combustion of CH4 in KJ mol−1

The heat of formations for $C 0_{2} (g), H_{2} 0 (1)$ and $C H_{4} (G)$ are $- 400 k J m o l^{- 1}, - 280 k J m o l^{- 1}$ and $- 70 k J m o l^{- 1}$ respectively. The heat of combustion of $C H_{4}$ in $m o l^{- 1}$ is?