Given are the following data at 298 K. ΔcH∘C2H4,g=−1410.9kJmol−1; ΔcH∘C2H6,g=−1559.8kJmol−1; ΔcH∘H2,g=−285.8kJmol−1 The value of ΔrH∘ of the reaction C2H4(g)+H2(g)⟶C2H6(g) will be

A
434.7 kJ mol^-1
B
-434.7 kJ mol^-1
C
136.9 kJ mol^-1
D
-136.9 kJ mol^-1

Solution:

$(i) C_{2} H_{4} (g) + 3 O_{2} (g) ⟶ 2 {CO}_{2} (g) + 2 H_{2} O (l); Δ_{c} H^{\circ} = - 1410.9 kJ {mol}^{- I}$

$(ii) C_{2} H_{6} (g) + \frac{7}{2} O_{2} (g) ⟶ 2 {CO}_{2} (g) + 3 H_{2} O; Δ_{c} H^{\circ} = - 1559.8 kJ {mol}^{- 1}$ $(iii) H_{2} (g) + \frac{1}{2} O_{2} (g) ⟶ H_{2} O (l);$ $Δ_{c} H^{\circ} = - 285.8 kJ {mol}^{- 1}$

The required chemical equation for which $Δ_{1} H^{\circ}$ ° is required is $C_{2} H_{4} (g) + H_{2} (g) ⟶ C_{2} H_{6} (g)$ This equation is obtained by the manipulations Eq.(i) + Eq. (iii) - Eq. (ii) Hence, $Δ_{r} H^{\circ} = [- 1410.9 - 285.8 - (- 1559.8)] kJ {mol}^{- 1} = - 136.9 kJ {mol}^{- 1}$

Solution:

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