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

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:  

$\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)]$

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:

$\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(CO_2)})+(2\times \Delta H^o_f_{(SO_2)})]-[(1\times \Delta H^o_f_{(CS_2)})+(3\times \Delta H^o_f_{(O_2)})]$

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:

$\Delta H^o_{rxn}=[(1\times (-393.52))+(2\times (-296.8))]-[(1\times (89.70))+(3\times (0)]\\\\\Delta H^o_{rxn}=-1076.82kJ$

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