Question

The chemistry of nitrogen oxides is very versatile. Given the following reactions and their standard enthalpy changes,

Answer 1

Answer:

The heat of the given reaction is -23.0 kJ.

Explanation:

We are given with ;

$NO(g) + NO_2(g)\rightarrow N_2O_3(g) \Delta H_{1,rxn}=-39.kJ$..[1]

$NO(g) + NO_2(g) + O_2(g)\rightarrow N_2O_5(g), \Delta H_{2,rxn} = -112.5 kJ$..[2]

$2NO_2(g) \rightarrow N_2O_4(g) ,\Delta H_{3,rxn} = -57.2 kJ$..[3]

$2NO(g) + O_2(g)\rightarrow 2NO_2(g), \Delta H_{4,rxn} = -114.2 kJ$..[4]

$N_2O_5(s)\rightarrow N_2O_5(g) ,\Delta H_{5,sub}= 54.1 kJ$..[5]

To find heat of reaction:

$N_2O_3(g) + N_2O_5(s)\rightarrow 2N_2O_4(g),\Delta H_{6,rxn} = ?$..[6]

Using Hess's law:

[5] +2 × [3] + [4] - [2] - [1] = [6]

$\Delta H_{6,rxn}=\Delta H_{5,sub}+2\times \Delta H_{3,rxn}+\Delta H_{4,rxn}-\Delta H_{2,rxn}-\Delta H_{1,rxn}$

$\Delta H_{6,rxn}=54.1 kJ+(2\times (-57.2 kJ))+(-114.2 kJ)-(-112.5 kJ)-(-39.0 kJ)$

$\Delta H_{6,rxn}=-23.0 kJ$

The heat of the given reaction is -23.0 kJ.