ΔH = +438 kJ
We have three equations:
(I) N₂ + 3H₂ → 2NH₃; Δ<em>H</em> = -92 kJ
(II) H₂ +½O₂ → H₂O; Δ<em>H</em> = -286 kJ
(III) CO(NH₂)₂ + ³/₂O₂ → CO₂ + 2H₂O + N₂; Δ<em>H</em> = -632 kJ
From these, we must devise the target equation:
(IV) 2NH₃ + CO₂ → CO(NH₂)₂ + H₂O; Δ<em>H</em> = ?
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The target equation has 2NH₃ on the left, so you <em>reverse equation (I)</em>.
When you reverse an equation, you <em>reverse the sign of its ΔH</em>.
(V) 2NH₃ → N₂ + 3H₂; Δ<em>H</em> = +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.
<em>Reverse Equation (III).</em>
(VI) CO₂ + 2H₂O + N₂ → CO(NH₂)₂ + ³/₂O₂; Δ<em>H</em> = +632 kJ
Equation <em>(VI)</em> has ³/₂O₂ on the right, and that is not in the target equation.
You need ³/₂O₂ on the left.
Multiply <em>Equation (II) by three</em>.
When you multiply an equation by three, you <em>multiply its ΔH by thre</em>e.
(VII) 3H₂ +³/₂O₂ → 3H₂O; Δ<em>H</em> = -286 kJ
Now, you add equations (V), (VI), and (VII), <em>cancelling species</em> that appear on opposite sides of the reaction arrows.
When you add equations, you add their Δ<em>H</em> values.
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We get the target equation (IV):
(V) 2NH₃ → <u>N</u>₂ + <u>3H</u>₂; ΔH = + 92 kJ
(VI) CO₂ + <u>2H</u>₂<u>O</u> + <u>N</u>₂ → CO(NH₂)₂ + ³/₂<u>O</u>₂; ΔH = +632 kJ
(VII) <u>3H</u>₂ +³/₂<u>O</u>₂ → <u>3</u>H₂O; ΔH = -286 kJ
(IV) 2NH₃ + CO₂ → CO(NH₂)₂ + H₂O; ΔH = +438 kJ
= 0.6944 mmol Ca(OH)₂