Question

Calcule a variação da entalpia dessa reação ( 2 NH3 (g) ---> CO(NH2)2 (s) + H2O (L) ) a partir das seguintes equações termoquímicas :

Answer 1

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