1. Tardigrade
  2. Question
  3. Chemistry
  4. 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

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Q. Ethanol can undergo decomposition to form two sets of products
$C _{2} H _{5} OH ( g ) \longrightarrow C _{2} H _{4}( g )+ H _{2} O ( g ) ; \,\,\,\, \Delta_{ r } H ^{\circ}=45.54 kJ$
$C _{2} H _{5} OH ( g ) \longrightarrow CH _{3} CHO ( g )+ H _{2}( g ) ; \,\,\,\, \Delta_{ r } H ^{\circ}=68.91 Kj$
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

Thermodynamics

Solution:

$C _{2} H _{4}$ and $CH _{3} CHO$ are in 8: 1 molar ratio

Thus, mole fraction $\Delta_{r} H^{\circ}$

$C _{2} H _{4} \therefore 8 / 9 \frac{8}{9} \times 45.54=40.48 \,kJ$

$CH _{3} CHO \therefore 1 / 9 \frac{1}{9} \times 68.91=7.66\, kJ$

Total $\Delta_{r} H^{\circ}=48.14\, kJ$