Question

use the table of standard thermodynamic properties as needed for the following questions.
hydrogen gas, h₂, reacts explosively with gaseous chlorine, cl₂, to form hydrogen chloride, hcl(g). what is the enthalpy change for the reaction of 1 mole of h₂(g) with 1 mole of cl₂(g) if both the reactants and products are at standard state conditions? the standard enthalpy of formation of hcl(g) is -92.3 kj/mol.
-46.2 kj
46.2 kj
184.6 kj
-184.6 kj
choose the correct heat of formation reaction for nano₃(s).
na(s) + ½n₂(g) + ³/₂o₂(g) → nano₃(s)
2na(s) + n₂(g) + 3o₂(g) → 2nano₃(s)
na(s) + n₂(g) + o₂(g) → nano₃(s)
na(s) + ½n₂(l) + ³/₂o₂(l) → nano₃(s)

Explanation:

Step1: Write the balanced chemical equation

The reaction is \(H_{2}(g)+Cl_{2}(g)\to2HCl(g)\)

Step2: Use the formula for enthalpy change of reaction \(\Delta H_{rxn}=\sum n\Delta H_{f(products)}-\sum m\Delta H_{f(reactants)}\)

For elements in their standard state, \(\Delta H_{f}=0\). So \(\Delta H_{f}(H_{2}) = 0\) and \(\Delta H_{f}(Cl_{2})=0\).
\(\Delta H_{rxn}=(2\times\Delta H_{f}(HCl))-(1\times\Delta H_{f}(H_{2}) + 1\times\Delta H_{f}(Cl_{2}))\)
Substitute \(\Delta H_{f}(HCl)=- 92.3\space kJ/mol\), \(\Delta H_{f}(H_{2}) = 0\), \(\Delta H_{f}(Cl_{2})=0\)
\(\Delta H_{rxn}=(2\times(-92.3))-(0 + 0)\)
\(\Delta H_{rxn}=-184.6\space kJ\)

The standard heat of formation (\(\Delta H_{f}^{\circ}\)) of a compound is the enthalpy change when 1 mole of the compound is formed from its elements in their standard states. For the first reaction \(H_{2}(g)+Cl_{2}(g)\to2HCl(g)\), using \(\Delta H_{rxn}=\sum n\Delta H_{f(products)}-\sum m\Delta H_{f(reactants)}\) with \(\Delta H_{f}(H_{2})=\Delta H_{f}(Cl_{2}) = 0\) (elements in standard state) and \(\Delta H_{f}(HCl)=-92.3\space kJ/mol\), we get \(\Delta H_{rxn}=2\times(-92.3)-0=-184.6\space kJ\).

For the second part (heat of formation of \(NaNO_{3}(s)\)):
The standard heat of formation reaction for a compound has 1 mole of the compound as the product and elements in their standard states as reactants.

• Option 1: \(Na(s)+\frac{1}{2}N_{2}(g)+\frac{3}{2}O_{2}(g)\to NaNO_{3}(s)\) has 1 mole of \(NaNO_{3}(s)\) as product and \(Na(s)\), \(N_{2}(g)\), \(O_{2}(g)\) (standard states of elements) as reactants.
• Option 2: Has 2 moles of \(NaNO_{3}(s)\) as product (not 1 mole as required for \(\Delta H_{f}\)).
• Option 3: The stoichiometry of elements is not balanced to form \(NaNO_{3}\) from standard - state elements.
• Option 4: \(N_{2}(l)\) and \(O_{2}(l)\) are not the standard states (standard state of \(N_{2}\) and \(O_{2}\) is gas at \(25^{\circ}C\) and 1 atm).

Answer:

-184.6 kJ