Ethanol can undergo decomposition to form two sets of products C 2 H 5 OH ( g ) longrightarrow C 2 H 4( g )+ H 2 O ( g ) ; Δ r H °=45.54 kJ C 2 H 5 OH ( g ) longrightarrow CH 3 CHO ( g )+ H 2( g ) ; Δ r H °=68.91 Kj If the molar ratio of C 2 H 4 to CH 3 CHO is 8: 1 in a set of product gases, then, energy involved in the decomposition process is

Question

Ethanol can undergo decomposition to form two sets of products C 2 H 5 OH ( g ) longrightarrow C 2 H 4( g )+ H 2 O ( g ) ; Δ r H °=45.54 kJ C 2 H 5 OH ( g ) longrightarrow CH 3 CHO ( g )+ H 2( g ) ; Δ r H °=68.91 Kj If the molar ratio of C 2 H 4 to CH 3 CHO is 8: 1 in a set of product gases, then, energy involved in the decomposition process is (A) 14.31 kJ (B) 57.23 kJ (C) 48.14 kJ (D) 66.31 kJ

Tardigrade Admin · Accepted Answer

C 2 H 4 and CH 3 CHO are in 8: 1 molar ratio Thus, mole fraction Δr H° C 2 H 4 ∴ 8 / 9 (8/9) × 45.54=40.48 kJ CH 3 CHO ∴ 1 / 9 (1/9) × 68.91=7.66 kJ Total Δr H°=48.14 kJ

14.31 kJ

57.23 kJ

48.14 kJ

66.31 kJ

C2​H4​ and CH3​CHO are in 8: 1 molar ratio Thus, mole fraction Δr​H∘ C2​H4​∴8/998​×45.54=40.48kJ CH3​CHO∴1/991​×68.91=7.66kJ Total Δr​H∘=48.14kJ

$C_{2} H_{4}$ and $C H_{3} C H O$ are in 8: 1 molar ratio

Thus, mole fraction $Δ_{r} H^{\circ}$

$C_{2} H_{4} ∴ 8/9 \frac{8}{9} \times 45.54 = 40.48 k J$

$C H_{3} C H O ∴ 1/9 \frac{1}{9} \times 68.91 = 7.66 k J$

Total $Δ_{r} H^{\circ} = 48.14 k J$