Question-50 Calculate the standard enthalpy change of following reaction.
C_{2}H_{4(g)} + 3O_{2(g)} \rightarrow 2CO_{2(g)} + 2H_{2}O_{(l)}If \Delta_{f}H^{\circ}(C_{2}H_{4}) = -52\ kJ\ mol^{-1}
\Delta_{f}H^{\circ}(CO_{2}) = -390\ kJ\ mol^{-1} \Delta_{f}H^{\circ}(H_{2}O) = -286\ kJ\ mol^{-1}(A) -650 kJ
(B) -1300 kJ
(C) -1950 kJ
(D) -1500 kJ
Answer: (B)
Explanation:
\Delta_r H^\circ=\sum \Delta_f H^\circ_{(products)}-\sum \Delta_f H^\circ_{(reactants)}\Delta_f H^\circ_{(O_2)}=0, because it is in its standard elemental state.
\Delta_r H^\circ=2\Delta_f H^\circ_{(CO_2)}+2\Delta_f H^\circ_{(H_2O)}-\Delta_f H^\circ_{(C_2H_4)}= 2(−390) + 2(−286) − (−52)
= −780 − 572 − (−52)
= −780 − 572 + 52 = −1300 kJ