Question

PLZ HELP For the reaction: 2NO2(g) → N2O4(l),

the ΔH of the reactants, two moles of NO2 (g), is + 66 kJ/mol,
and the ΔH of the products, N2O4 (l), is -20 kJ/mol.
Which of the following shows the ΔH (change in enthalpy) for the reaction as a whole?

Question 8 options:

ΔrxnH =(- 20 kJ/mol) / (+66 kJ/mol)


ΔrxnH = (+66 kJ/mol) + (- 20 kJ/mol)


ΔrxnH =(-20 kJ/mol) - (+ 66 kJ/mol)


ΔrxnH =(+66 kJ/mol) - (- 20 kJ/mol)

Answer 1

Answer:
`\Delta H_(rxn)=-20kJ/mol-(+66kJ/mol)`

Step-by-step explanation:

Heat of reaction or enthalpy change is the energy released or absorbed during the course of the reaction.

It is calculated by subtracting the enthalpy of reactants from the enthalpy of products.

`\Delta H=H_(products)-H_(reactants)`

`\Delta H` = enthalpy change = ?

`H_(products)` = enthalpy of products

`H_(reactants)` = enthalpy of reactants

For the given reaction :

`2NO_2(g)\rightarrow N_2O_4(l)`

`\Delta H=H_(N_2O_4)-2* H_(NO_2)`

`\Delta H=-20kJ/mol-(+66kJ/mol)`