Question

The ∆ f H° for CO₂(g) CO(g) and H₂O(g) are –393.5, –110.5
and –241.8 kJ/mol respectively, the standard enthalpy change
(in kJ) for the reaction CO₂(g) + H₂(g) → CO(g) + H₂O(g) is :
(a) 524.1 (b) 41.2
(c) – 262.5 (d) – 41.2

Answer 1

The standard enthalpy change of the reaction is 41.2.

Step by step explanation:

"Standard heat for the formation of a compound is the change in enthalpy from its constituent elements during the formation of 1 mole of the substance."

From the given,

$\Delta f\,H^{o}\,of\,CO_{2}=-393.5$

$\Delta f\,H^{o}\,of\,CO =-110.5$

$\Delta f\,H_{o}\,of\,H_{2}O =-241.8$

        $\bold{\Delta\,H_{reaction}^{o}=\Delta\,H_{f}^{o}(CO)+\Delta\,H_{f}^{o}(H_{2}O)-\Delta\,H_f^{o}(CO_{2})}$

Substitute the given values.

                           $\Rightarrow -110.5-241.8+393.5$

                           $\Rightarrow 41.2$

Therefore, The standard enthalpy change of the reaction is 41.2kj/mol.

Hence, correct option is "b".

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Answer 2

(b) 41.2

Explanation:

Standard heat for the formation of a compound is the change in enthalpy from its constituent elements during the formation of 1 mole of the substance.

Change in enthalpy of CO2 = -393.5

Change in enthalpy of CO = -110.5

Change in enthalpy of H2O = -241.8

Change in enthalpy of the reaction = Change in enthalpy of CO + Change in enthalpy of water - Change in enthalpy of CO2  

= -110.5 -241.8 + 393.5 = 41.2

The standard enthalpy change of the given reaction is 41.2 kj/mol.

Option B is the answer.