The standard enthalpies of combustion (to CO2(g) and H2O(l)) are -1300 for C2H2(g), -1411 for C2H4(g) and -286 for H2(g), all in kJ mol^-1. Calculate Delta H for C2H2(g) + H2(g) -> C2H4(g) in kJ mol^-1 (nearest integer).

JEE Main levelNumerical answerDifficulty 2 of 3td131

The standard enthalpies of combustion (to $CO X_{2} (g)$ and $H X_{2} O (l)$ ) are $- 1300$ for $C X_{2} H X_{2} (g)$ , $- 1411$ for $C X_{2} H X_{4} (g)$ and $- 286$ for $H X_{2} (g)$ , all in $kJ mo l^{- 1}$ . Calculate $Δ H$ for $C X_{2} H X_{2} (g) + H X_{2} (g) C X_{2} H X_{4} (g)$ in $kJ mo l^{- 1}$ (nearest integer).

Numerical answer type, answer in kJ/mol, tolerance 0.5.

Hess's law:

Δ H = \sum Δ_{c} H (reactants) - \sum Δ_{c} H (products) = [(- 1300) + (- 286)] - (- 1411) = - 1586 + 1411 = - 175 kJ mo l^{- 1}