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 = ?

Respuesta :

Answer:

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

Explanation:

Given:

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

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

To find ; [tex]\Delta H_{rxn}[/tex] of following reaction :

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

Using Hess's Law:

[2] - [1] = [3]

[tex]\Delta H_{rxn}=\Delta H_2-\Delta H_1=-395 kJ/mol-(-111 kJ/mol)=-284 kJ/mol[/tex]

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

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