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Determine the enthalpy:
4NH3(g) + 7O2(g) −→ 4NO2(g) + 6H2O(ℓ)

∆H◦ f species ( kJ mol)
NH3(g) −46.11
NO2(g) +33.2
H2O(ℓ) −285.83

1. −1899 kJ/mol rxn
2. +2034.4 kJ/mol rxn
3. −298.7 kJ/mol rxn
4. −1766 kJ/mol rxn
5. −1397.6 kJ/mol rxn

Respuesta :

jacob193

Answer:

Choice 5.: -1397.6 kJ/mol rxn.

Explanation:

Here's how to find the enthalpy change of a reaction [tex]\Delta H_\text{rxn}[/tex] from the standard enthalpy of formation [tex]\Delta H_\text{f}\textdegree{}[/tex] of each species.

  • Find the sum of [tex]\Delta H_\text{f}\textdegree{}[/tex] for the products: [tex]\Sigma \left[\Delta H_f\textdegree{}(\text{Products})]\right[/tex]. Multiply the [tex]\Delta H_\text{f}\textdegree{}[/tex] of each product by its coefficient in the equation. Find the sum of those products.
  • Find the sum of enthalpies of formation [tex]\Delta H_\text{f}\textdegree{}[/tex] for the reactants: [tex]\Sigma \left[\Delta H_f\textdegree{}(\text{Reactants})]\right[/tex]. Similarly, multiply the [tex]\Delta H_\text{f}\textdegree{}[/tex] of each reactant by its coefficient in the equation. Find the sum of those products.
  • The enthalpy change of this reaction is the same as the difference between [tex]\Sigma \left[\Delta H_f\textdegree{}(\text{Products})]\right[/tex] and [tex]\Sigma \left[\Delta H_f\textdegree{}(\text{Reactants})]\right[/tex]. Note the order: product minus reactants. Keep all the signs as is in these calculations.

[tex]\Delta H_\text{f}\textdegree{} = 0[/tex] for the most stable allotrope of each elements under their standard state. For example,

  • [tex]\text{O}_2[/tex] (not [tex]\text{O}_3[/tex] ozone) is the most stable allotrope of oxygen under STP.
  • [tex]\text{O}_2[/tex] is a gas under STP.

As a result, [tex]\Delta H_\text{f}\textdegree{} = 0[/tex] for [tex]\text{O}_2\;(g)[/tex]. Note the gaseous state symbol. This value is implied such that it's not provided in the question.

For this reaction:

[tex]\Sigma \left[\Delta H_f\textdegree{}(\textbf{Products})]\right = 4 \times (+33.2) + 6\times (-285.83) = -1582.18\;\text{kJ}\cdot\text{mol}^{-1}[/tex].

[tex]\Sigma \left[\Delta H_f\textdegree{}(\textbf{Reactants})]\right = 4 \times (-46.11) + 7\times 0 = -184.44 \;\text{kJ}\cdot\text{mol}^{-1}[/tex].

Apply the equation:

[tex]\begin{aligned} \Delta H_\text{rxn} &= \Sigma \left[\Delta H_f\textdegree{}(\textbf{Products})]\right - \Sigma \left[\Delta H_f\textdegree{}(\textbf{Reactants})]\right\\ &= -1582.18\;\text{kJ}\cdot\text{mol}^{-1} - (-184.44 \;\text{kJ}\cdot\text{mol}^{-1})\\ &= -1397.74\;\text{kJ}\cdot\text{mol}^{-1}\end{aligned}[/tex].

pstnonsonjoku

The enthalpy change of the reaction −1397.6 kJ/mol.

We can determine the enthalpy change of the reaction from the enthalpy of formation of each specie as follows;

∆H(reaction) = ∑∆H◦f(products) - ∑∆H◦f(reactants)

Given the following from the question;

NH3(g) = −46.11 kJ /mol

NO2(g) = +33.2 kJ /mol

H2O(ℓ) = −285.83 kJ /mol

We can now substitute values;

∆H(reaction) =   ∑(4 × 33.2) + (6 × (−285.83)) - ∑(4 × (−46.11)) + (7 × 0))  kJ /mol

The ∆H◦ f  of oxygen is zero because it is a pure substance.

∆H(reaction) = (132.8 - 1714.98) - (-184.44 + 0) kJ /mol

∆H(reaction) =−1397.6 kJ/mol

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