Question

using the data below, calculate the enthalpy for the decomposition of no₂ into no and o₂. 2no₂(g) → 2no(g) + o₂(g) \

compound	s° (j/mol k)	δh°f (kj/mol)	\
---	---	---	\
no₂(g)	240	34	\
no(g)	211	90	\
o₂(g)	205	0	\

δhᵣₓₙ = ? kj \
enter either a + or - sign and the magnitude.

Explanation:

Step1: Recall the formula for enthalpy of reaction

The formula for the enthalpy of a reaction (\(\Delta H_{rxn}\)) is \(\Delta H_{rxn}=\sum n\Delta H_f^{\circ}(\text{products})-\sum m\Delta H_f^{\circ}(\text{reactants})\), where \(n\) and \(m\) are the stoichiometric coefficients of products and reactants respectively.

Step2: Identify products and reactants with coefficients

For the reaction \(2NO_2(g)
ightarrow 2NO(g)+O_2(g)\), the products are \(2\) moles of \(NO(g)\) and \(1\) mole of \(O_2(g)\), and the reactant is \(2\) moles of \(NO_2(g)\).

Step3: Calculate \(\sum n\Delta H_f^{\circ}(\text{products})\)

For \(NO(g)\): \(n = 2\), \(\Delta H_f^{\circ}=90\space kJ/mol\), so contribution is \(2\times90 = 180\space kJ\).
For \(O_2(g)\): \(n = 1\), \(\Delta H_f^{\circ}=0\space kJ/mol\), so contribution is \(1\times0 = 0\space kJ\).
Sum of products: \(180 + 0=180\space kJ\).

Step4: Calculate \(\sum m\Delta H_f^{\circ}(\text{reactants})\)

For \(NO_2(g)\): \(m = 2\), \(\Delta H_f^{\circ}=34\space kJ/mol\), so contribution is \(2\times34 = 68\space kJ\).

Step5: Calculate \(\Delta H_{rxn}\)

Using the formula \(\Delta H_{rxn}=\sum n\Delta H_f^{\circ}(\text{products})-\sum m\Delta H_f^{\circ}(\text{reactants})\)
\(\Delta H_{rxn}=180 - 68=112\space kJ\)

Answer:

\(+112\)