Calcule a variação da entalpia dessa reação ( 2 nh3 (g) > co(nh2)2 (s) + h2o (l) ) a partir das seguintes equações termoquímicas :

ΔH = +438 kJ  

We have three equations:  

(I) N₂ + 3H₂ → 2NH₃; ΔH = -92 kJ  

(II) H₂ +½O₂ → H₂O; ΔH = -286 kJ  

(III) CO(NH₂)₂ + ³/₂O₂ → CO₂ + 2H₂O + N₂; ΔH = -632 kJ  

From these, we must devise the target equation:  

(IV) 2NH₃ + CO₂ → CO(NH₂)₂ + H₂O; ΔH = ?  

The target equation has 2NH₃ on the left, so you reverse equation (I).  

When you reverse an equation, you reverse the sign of its ΔH.  

(V) 2NH₃ → N₂ + 3H₂; ΔH = +92 kJ  

Equation (V) has 1N₂ on the right, and that is not in the target equation.  

You need an equation with 1N₂ on the left.  

Reverse Equation (III).  

(VI) CO₂ + 2H₂O + N₂ → CO(NH₂)₂ + ³/₂O₂; ΔH = +632 kJ  

Equation (VI) has ³/₂O₂ on the right, and that is not in the target equation.  

You need ³/₂O₂ on the left.  

Multiply Equation (II) by three.  

When you multiply an equation by three, you multiply its ΔH by three.

(VII) 3H₂ +³/₂O₂ → 3H₂O; ΔH = -286 kJ  

Now, you add equations (V), (VI), and (VII), cancelling species that appear on opposite sides of the reaction arrows.  

When you add equations, you add their ΔH values.  

We get the target equation (IV):  

(V) 2NH₃ → N₂ + 3H₂;                                    ΔH = +  92 kJ  

(VI) CO₂ + 2H₂O + N₂ → CO(NH₂)₂ + ³/₂O₂; ΔH = +632 kJ  

(VII) 3H₂ +³/₂O₂ → 3H₂O;                             ΔH =   -286 kJ

(IV) 2NH₃ + CO₂ → CO(NH₂)₂ + H₂O;          ΔH =  +438 kJ  

since electrons are negative the answer is neagtaive