Question

Carbon disulfide is a colorless liquid. When pure, it is nearly odorless, but the commercial product smells vile. Carbon disulfide is used in the manufacture of rayon and cellophane. The liquid burns as follows: CS2(l) + 3O2(g) → CO2(g) + 2SO2(g)Calculate the standard enthalpy change for this reaction usingstandard enthalpies of formation.

Answer 1

Answer: The standard enthalpy change of the reaction is -1076.82kJ

Step-by-step explanation:

Enthalpy change 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
`\Delta H^o`

The equation used to calculate enthalpy change is of a reaction is:

`\Delta H^o_(rxn)=\sum [n* \Delta H^o_f(product)]-\sum [n* \Delta H^o_f(reactant)]`

For the given chemical reaction:

`CS_2(l)+3O_2(g)\rightarrow CO_2(g)+2SO_2(g)`

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

`\Delta H^o_(rxn)=[(1* \Delta H^o_f_((CO_2)))+(2* \Delta H^o_f_((SO_2)))]-[(1* \Delta H^o_f_((CS_2)))+(3* \Delta H^o_f_((O_2)))]`

We are given:

`\Delta H^o_f_((CO_2))=-393.52kJ/mol\\\Delta H^o_(SO_2)=-296.8kJ/mol\\\Delta H^o_f_((O_2))=0kJ/mol\\\Delta H^o_(CS_2)=89.70kJ/mol`

Putting values in above equation, we get:

`\Delta H^o_(rxn)=[(1* (-393.52))+(2* (-296.8))]-[(1* (89.70))+(3* (0)]\\\\\Delta H^o_(rxn)=-1076.82kJ`

Hence, the standard enthalpy change of the reaction is -1076.82kJ