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murrachl000
05.05.2020 •
Chemistry
What is the mass in grams of .802mol of salycic acid
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Ответ:
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