nitrogen dioxide, no2(g) (δhf = 33.84 kj/mol), is decomposed according to the following reaction: 2no2(g) →…

Explanation:

Step1: Determine $\Delta H_f$ values for products

The standard - enthalpy of formation of $N_2(g)$ is $\Delta H_f(N_2)=0$ kJ/mol and for $O_2(g)$ is $\Delta H_f(O_2)=0$ kJ/mol.

Step2: Calculate $\sum(\Delta H_{f,products})$

For the reaction $2NO_2(g)\rightarrow N_2(g) + 2O_2(g)$, $\sum(\Delta H_{f,products})=\Delta H_f(N_2)+2\Delta H_f(O_2)=0 + 2\times0=0$ kJ/mol.

Step3: Calculate $\sum(\Delta H_{f,reactants})$

The $\Delta H_f$ for $NO_2$ is given as $\Delta H_f(NO_2)=33.84$ kJ/mol. For 2 moles of $NO_2$, $\sum(\Delta H_{f,reactants}) = 2\times\Delta H_f(NO_2)=2\times33.84 = 67.68$ kJ/mol.

Step4: Calculate $\Delta H_{rxn}$

Using the formula $\Delta H_{rxn}=\sum(\Delta H_{f,products})-\sum(\Delta H_{f,reactants})$, we get $\Delta H_{rxn}=0 - 67.68=- 67.68$ kJ for the reaction of 2 moles of $NO_2$.

Step5: Calculate $\Delta H$ for 2.50 mol of $NO_2$

Set up a proportion. If for 2 moles $\Delta H=-67.68$ kJ, for 2.50 moles, $\Delta H=\frac{2.50}{2}\times67.68 = 84.6$ kJ. Since $\Delta H>0$, energy is absorbed.

Answer:

84.6 kJ of energy absorbed