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Yuki888 [10]
3 years ago
5

Calculate the heat of decomposition for this process at constant pressure and 25°C: CaCO3(s) → CaO(s) + CO2(g) The standard enth

alpies of formation for the reactant are: ΔHf CaCO3(s) = −1206.9 kJ/mol ; ΔHf CaO(s) = −635.6 kJ/mol; ΔHf CO2(g) −393.5 kJ/mol
Chemistry
1 answer:
ololo11 [35]3 years ago
8 0

Answer:

177.8kJ/mol

Explanation:

In this reaction, the heat of decomposition is the same as the heat of formation. This is a decomposition reaction.

Given parameters:

ΔHf CaCO₃ = -1206.9kJ/mol

ΔHf CaO = −635.6 kJ/mol

ΔHf CO₂ = −393.5 kJ/mol

The heat of decomposition =

                     Sum of ΔHf of products - Sum of ΔHf of reactants

The equation of the reaction is shown below:

     CaCO₃ → CaO + CO₂

The heat of decomposition = [ -635.6 + (-393.5)] - [−1206.9 ]

                                             = -1029.1 + 1206.9

                                             = 177.8kJ/mol

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Answer: 2 moles of H_2O will be formed.

Explanation:

To calculate the moles :

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The balanced chemical equation is:

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According to stoichiometry :

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Answer:

The rate of consumption of NH_{3} is 2.0 mol/L.s

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where -\frac{\Delta [NH_{3}]}{\Delta t} represents rate of consumption of NH_{3}, -\frac{\Delta [O_{2}]}{\Delta t} represents rate of consumption of O_{2}, \frac{\Delta [N_{2}]}{\Delta t} represents rate of formation of N_{2} and \frac{\Delta [H_{2}O]}{\Delta t} represents rate of formation of H_{2}O.

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From the above equation we can write-

-\frac{1}{4}\frac{\Delta [NH_{3}]}{\Delta t}=\frac{1}{6}\frac{\Delta [H_{2}O]}{\Delta t}

Here \frac{\Delta [H_{2}O]}{\Delta t}=3.0 mol/(L.s))

So, -\frac{\Delta [NH_{3}]}{\Delta t}=\frac{4}{6}\frac{\Delta [H_{2}O]}{\Delta t}

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