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Firlakuza [10]
3 years ago
8

Calculate the DH°rxn for the decomposition of calcium carbonate to calcium oxide and carbon dioxide. DH°f means delta or change

of heat of formation DH°f [CaCO3(s)] = –1206.9 kJ/mol; DH°f [CaO(s)] = –635.1 kJ/mol; DH°f [CO2(g)] = –393.5 kJ/mol CaCO3(s) --> CaO(s) + CO2(g)
Chemistry
1 answer:
USPshnik [31]3 years ago
5 0

Answer: +178.3 kJ

Explanation:

The chemical equation  follows:

CaCO_3(s)\rightarrow CaO(s)+CO_2(g)

The equation for the enthalpy change of the above reaction is:

\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(CaO(s))})+(1\times \Delta H^0f_{CO_2}]-[(1\times \Delta H^o_f_{(CaCO_3(s))})]

We are given:

\Delta H^o_f_{(CaO(s))}=-635.1kJ/mol\\\Delta H^o_f_{(CaCO_3(s))}=-1206.9kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_{rxn}=?

Putting values in above equation, we get:

\Delta H^o_{rxn}=[(1\times (-635.1))+(1\times (-393.5))]-[(1\times (-1206.9))]

The DH°rxn for the decomposition of calcium carbonate to calcium oxide and carbon dioxide is +178.3 kJ

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Answer is: ph value is 3.56.
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Chemical reaction 2: HCO₃⁻(aq) ⇄ CO₃²⁻(aq) + H⁺(aq); Ka₂ = 5,6·10⁻¹¹.
c(H₂CO₃) = 0,18 M.
[HCO₃⁻] = [H⁺<span>] = x.
</span>[H₂CO₃] = 0,18 M - x.
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