Question

ethanol, c₂h₅oh, is made industrially by the reaction of water with ethylene, c₂h₄. calculate the value of δᵣh° for the reaction given the following thermochemical equations. c₂h₄(g)+h₂o(l)→c₂h₅oh(l) c₂h₄(g)+3o₂(g)→2co₂(g)+2h₂o(l) δᵣh° = -1411.1 kj c₂h₅oh(l)+3o₂(g)→2co₂(g)+3h₂o(l) δᵣh° = -1367.1 kj (show all your work directly below, type it in using this mathtype application.)

Explanation:

Step1: Write the target reaction and given reactions

The target reaction is \(C_2H_4(g)+H_2O(l)\longrightarrow C_2H_5OH(l)\).
Given reaction 1: \(C_2H_4(g)+3O_2(g)\longrightarrow 2CO_2(g)+2H_2O(l)\), \(\Delta H_1=- 1411.1\ kJ\).
Given reaction 2: \(C_2H_5OH(l)+3O_2(g)\longrightarrow 2CO_2(g)+3H_2O(l)\), \(\Delta H_2=-1367.1\ kJ\).

Step2: Manipulate the given - reactions

We want to cancel out \(CO_2\) and \(O_2\) to get the target reaction.
Keep reaction 1 as it is: \(C_2H_4(g)+3O_2(g)\longrightarrow 2CO_2(g)+2H_2O(l)\), \(\Delta H_1=-1411.1\ kJ\).
Reverse reaction 2: \(2CO_2(g)+3H_2O(l)\longrightarrow C_2H_5OH(l)+3O_2(g)\), \(\Delta H_3 = 1367.1\ kJ\) (when a reaction is reversed, the sign of \(\Delta H\) changes).

Step3: Add the manipulated reactions

Add the two - manipulated reactions:
\((C_2H_4(g)+3O_2(g)\longrightarrow 2CO_2(g)+2H_2O(l))+(2CO_2(g)+3H_2O(l)\longrightarrow C_2H_5OH(l)+3O_2(g))\)
The \(3O_2(g)\) and \(2CO_2(g)\) on both sides cancel out, and we get \(C_2H_4(g)+H_2O(l)\longrightarrow C_2H_5OH(l)\).
According to Hess's law, \(\Delta H=\Delta H_1+\Delta H_3\).
\(\Delta H=-1411.1\ kJ + 1367.1\ kJ=-44\ kJ\).

Answer:

\(-44\ kJ\)