Question #50c2f

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Question #50c2f

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Answer 1 · 2015-03-24T22:00:08

The standard heat of reaction is -178.3 kJ.

We have the following information:

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For most chemistry problems involving $Δ H_{f}^{o}$ $ΔH_f^"o"$ , you need the following equation:

$Δ H_{r x n}^{o} = Σ Δ H_{f}^{o} (p) - Σ Δ H_{f}^{o} (r)$ $ΔH_(rxn)^"o" = ΣΔH_f^"o"("p") - ΣΔH_f^"o"("r")$ ,

where $p$ $"p"$ = products and $r$ $"r"$ = reactants.

$Σ Δ H_{f}^{o} (p) = 1 mol CaCO₃ \times \frac{-1206.9 kJ}{1 mol CaCO₃} = -1206.9 kJ$ $ΣΔH_f^"o"("p") = 1 cancel("mol CaCO₃") × "-1206.9 kJ"/(1 cancel("mol CaCO₃")) = "-1206.9 kJ"$

$Σ Δ H_{f}^{o} (r) = 1 mol CaO \times \frac{-635.09 kJ}{1 mol CaO} + 1 mol CO₂ \times \frac{-393.509 kJ}{1 mol CO₂} = -1028.599 kJ$ $ΣΔH_f^"o"("r") = 1 cancel("mol CaO") × "-635.09 kJ"/(1cancel("mol CaO")) + "1 mol" cancel("CO₂") × "-393.509 kJ"/(1cancel("mol CO₂")) = "-1028.599 kJ"$

$Δ H_{rxn}^{o} = Σ Δ H_{f}^{o} (p) - Σ Δ H_{f}^{o} (r) = -1206.9 kJ + 1028.599 kJ = -178.3 kJ$ $ΔH_("rxn")^"o" = ΣΔH_f^"o"("p") - ΣΔH_f^"o"("r") = "-1206.9 kJ + 1028.599 kJ" = "-178.3 kJ"$

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Question #50c2f

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