Question

Given the following thermochemical data C(s) + ½O2(g) ---> CO(g) ∆H1 = -111 kJ/mol C(s) + O2(g) ---> CO2(g) ∆H2 = -395 kJ/mol What is the ∆Hr for the reaction: CO(g) + ½O2(g) ---> CO2(g) ∆Hr = ?

Answer 1

The enthalpy of the reaction asked is -284 kJ/mol.

Step-by-step explanation:

Given:

`C(s) + (1)/(2)O_2(g)\rightarrow CO(g), \Delta H_1 = -111 kJ/mol`..[1]

`C(s) + O_2(g)\rightarrow CO_2(g) \Delta H_2 = -395 kJ/mol`...[2]

To find ;
`\Delta H_(rxn)` of following reaction :

`CO(g) + (1)/(2)O_2(g)\rightarrow CO_2(g), \Delta H_(rxn) =?`..[3]

Using Hess's Law:

[2] - [1] = [3]

`\Delta H_(rxn)=\Delta H_2-\Delta H_1=-395 kJ/mol-(-111 kJ/mol)=-284 kJ/mol`

The enthalpy of the reaction asked is -284 kJ/mol.