ethanol, c₂h₅oh, is made industrially by the reaction of water with ethylene, c₂h₄. calculate the value of…

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.)

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.)

Answer

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)