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Question: For a reaction, \[CaCO_{3(s)} \rightarrow CaO_{(s)} + CO_{2(s)}\] \[\Delta_{f}H^{o}(CaO) = - 635.1...

For a reaction,

CaCO3(s)→CaO(s)+CO2(s)CaCO_{3(s)} \rightarrow CaO_{(s)} + CO_{2(s)}

ΔfHo(CaO)=−635.1kJmol−1,\Delta_{f}H^{o}(CaO) = - 635.1kJmol^{- 1},

ΔfHo(CO2)=−393.5kJmol−1and\Delta_{f}H^{o}(CO_{2}) = - 393.5kJmol^{- 1}and

ΔfHo(CaCO3)=−1206.9kJmol−1\Delta_{f}H^{o}(CaCO_{3}) = - 1206.9kJmol^{- 1}

Which of the following is a correct statement?

A

A large amount of heat is evolved during the decomposition of CaCO3CaCO_{3}

B

Decomposition of CaCO3CaCO_{3} is an endothermic process and heat is provided for decomposition.

C

The amount of heat evolved cannot be calculated from the data provided.

D

ΔrHo=∑ΔrHo\Delta_{r}H^{o} = \sum\Delta_{r}H^{o} (reactants)−∑ΔrHo- \sum\Delta_{r}H^{o} (products)

Answer

Decomposition of CaCO3CaCO_{3} is an endothermic process and heat is provided for decomposition.

Explanation

Solution

: ΔrH=∑fHPo−∑ΔfHRo\Delta_{r}H = \sum_{f}H_{P}^{o} - \sum\Delta_{f}H_{R}^{o}

=[−635.1+(−393.5)]−[−1206.9]=+178.4kJ= \left\lbrack - 635.1 + ( - 393.5) \right\rbrack - \lbrack - 1206.9\rbrack = + 178.4kJ