Answer : The standard enthalpy of the reaction is 20.2 kJ
Explanation :
Enthalpy change : It is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as [tex]\Delta H^o[/tex]
The equation used to calculate enthalpy change is of a reaction is:
[tex]\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)][/tex]
The equilibrium reaction follows:
[tex]CS_2(g)+2H_2O(l)\rightleftharpoons CO_2(g)+2H_2S(g)[/tex]
The equation for the enthalpy change of the above reaction is:
[tex]\Delta H^o_{rxn}=[n_{(CO_2)}\times \Delta H^o_f_{(CO_2)}+n_{(H_2S)}\times \Delta H^o_f_{(H_2S)}]-[n_{(H_2O)}\times \Delta H^o_f_{(H_2O)}+n_{(CS_2)}\times \Delta H^o_f_{(CS_2)}][/tex]
We are given:
[tex]\Delta H^o_f_{(CS_2(g))}=116.7kJ/mol\\\Delta H^o_f_{(H_2O(l))}=-285.8kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_f_{(H_2S(g))}=-20.60kJ/mol[/tex]
Putting values in above equation, we get:
[tex]\Delta H^o_{rxn}=[(1mol\times -393.5kJ/mol)+(2mol\times -20.60kJ/mol)]-[(1mol\times 116.7kJ/mol)+(2mol\times -285.8kJ/mol)]=20.2kJ[/tex]
Therefore, the standard enthalpy of the reaction is 20.2 kJ
The standard enthalpy of reaction is 20.2 kJ/mol.
We can calculate the standard enthalpy of reaction from the standard heat of formation using the formula;
ΔHreaction = ∑ΔH∘f products - ΔH∘f reactants
The equation of the reaction is; CS2(g)+2H2O(l)→CO2(g)+2H2S(g)
ΔH∘f CS2 = 116.7 KJ/mol
ΔH∘f H2O(l) = −285.8 KJ/mol
ΔH∘f CO2 = −393.5 kJ/mol
ΔH∘f H2S= −20.60 kJ/mol
Substituting values, we have;
ΔHreaction = ∑[(−393.5) + 2 ×(−20.60) - (116.7) + 2 ×(−285.8)] kJ/mol
ΔHreaction = 20.2 kJ/mol
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