Question

Calculate the standard reaction enthalpy for the reaction NO2(g) → NO(g) + O(g) given +142.7 kJ/mol for the standard enthalpy of formation of ozone and O2(g) → 2 O(g) ∆H ◦ = +498.4 kJ/mol NO(g) + O3(g) → NO2(g) + O2(g) ∆H◦ = −200 kJ/mol Remember the definition of the standard enthalpy of formation of a substance.

Answer 1

The standard reaction enthalpy for the given reaction is 235.15 kJ/mol.

Step-by-step explanation:

`O_2(g) \rightarrow (2)/(3)O_3(g),\Delta H^o_(1)=142.7 kJ/mol`..[1]

`O_2(g) \rightarrow 2 O(g),\Delta H^o_(2)=498.4 kJ/mol`..[2]

`NO(g) + O_3(g)\rightarrow NO_2(g) + O_2(g) ,\Delta H^o_(3) = -200 kJ/mol`..[3]

`NO_2(g)\rightarrow NO(g) + O(g),\Delta H^o_(4)=?`..[4]

Using Hess's law:

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

2 × [4] = [2]- (3 ) × [1] - (2) × [3]

`2* \Delta H^o_(4)=\Delta H^o_(2) -3* \Delta H^o_(1)-2* \Delta H^o_(3)`

`2* \Delta H^o_(4)=498.4 kJ/mol-3* 142.7 kJ/mol-2* -200 kJ/mol`

`2* \Delta H^o_(4)=470.3 kJ/mol`

`\Delta H^o_(4)=(470.3 kJ/mol)/(2)=235.15 kJ/mol`

The standard reaction enthalpy for the given reaction is 235.15 kJ/mol.