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Heat of Formation

The combustio...

Question

The combustion enthalpies of carbon, hydrogen and ethyne are $- 393.5 k J m o l^{- 1},$ $- 285.8 k J m o l^{- 1}$ and $- 1309.5 k J m o l^{- 1}$ respectively at $25^{o} C$ .
The value of standard enthalpy of formation of acetylene at this temperature is:

236.2 k J m o l^{- 1}

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298.8 k J m o l^{- 1}

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74.7 k J m o l^{- 1}

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- 107.7 k J m o l^{- 1}

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Solution

The correct option is C $236.2 k J m o l^{- 1}$
Combustion enthalpy of Carbon-
$C + O_{2} \to C O_{2}$ (i)
Combustion enthalpy of Hydrogen-
$H_{2} + \frac{1}{2} O_{2} \to H_{2} O$ (ii)
Comustion enthalpy of ethyne-
$C H \equiv C H + O_{2} \to 2 C O_{2} + H_{2} O$ (iii)
Formation enthalpy of acetylene-
$2 C + 2 H (g) \to H C \equiv C H$
To get the above reaction, multiply the 1st equation with $2$ and reverse the third reaction,
$Δ H_{f} = - (2 \times 393.5 + 285.8) + 1309.5 K c a l$
$Δ H_{f} = 236.2 \frac{k J}{m o l}$

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The combustion enthalpies of carbon, hydrogen and ethyne are −393.5 kJ mol−1, −285.8 kJ mol−1 and −1309.5 kJ mol−1 respectively at 25oC. The value of standard enthalpy of formation of acetylene at this temperature is:

The combustion enthalpies of carbon, hydrogen and ethyne are $- 393.5 k J m o l^{- 1},$ $- 285.8 k J m o l^{- 1}$ and $- 1309.5 k J m o l^{- 1}$ respectively at $25^{o} C$ .
The value of standard enthalpy of formation of acetylene at this temperature is: