Hess law of constant summation

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Question

Standard enthalpies of formation of ${CO}_{2}, H_{2} O and C_{2} H_{4}$ are -396 kJ/mole - 288 kJ/mole and +52 kJ/ mole respectively. Enthalpy of combustion of Ethene is

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Solution

Given
$\begin{array}{l} C (s) + O_{2} (g) \overset{}{\to} {CO}_{2} (g); ΔH = - 396 kJ \to (1) \\ H_{2} (g) + \frac{1}{2} O_{2} (g) \overset{}{\to} H_{2} O (g); ΔH = - 288 kJ \to (2) \\ 2 C (s) + 2 H_{2} (g) \overset{}{\to} C_{2} H_{4} (g); ΔH = + 52 kJ \to (3) \end{array}$
Required
$C_{2} H_{4} (g) + 3 O_{2} (g) \overset{}{\to} 2 {CO}_{2} (g) + 2 H_{2} O (l); ΔH = ?$
$2 \times Eq (1) + 2 Eq (2) - Eq (3)$
$ΔH = 2 (- 396) + 2 (- 288) - (+ 52)$
$= - 792 - 576 - 52$
$= - 1420 kJ / mole$

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Hess law of constant summation

Unlock the full solution & master the concept.

Standard enthalpies of formation of CO2, H2O and C2H4 are -396 kJ/mole - 288 kJ/mole and +52 kJ/ mole respectively. Enthalpy of combustion of Ethene is

GivenCs+O2g→ CO2g;ΔH=−396kJ→1H2g+12O2g→ H2Og;ΔH=−288kJ→22Cs+2H2g→ C2H4g;ΔH=+52kJ→3RequiredC2H4g+3O2g→ 2CO2g+2H2Ol;ΔH=?2×Eq1+2Eq2−Eq3ΔH=2−396+2−288−+52=−792−576−52=−1420kJ/mole

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Standard enthalpies of formation of ${CO}_{2}, H_{2} O and C_{2} H_{4}$ are -396 kJ/mole - 288 kJ/mole and +52 kJ/ mole respectively. Enthalpy of combustion of Ethene is