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  • Consider the reaction: <img alt="\mathrm{2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{NO}_2(\mathrm{~g})}" src="https://entrancecorner.oncodecogs.com/gif.late

Consider the reaction:

\mathrm{2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{NO}_2(\mathrm{~g})}The standard enthalpy change\mathrm{\left(\Delta H^{\circ}\right)} for this reaction is −113 kJ/mol. Use this information to determine the enthalpy change for the reaction when 3 moles of NO(g) react with excess \mathrm{\mathrm{O}_2(\mathrm{~g}) \text { to form } \mathrm{NO}_2(\mathrm{~g}) \text {. }}

Option: 1

165 kJ


Option: 2

169.5 kJ


Option: 3

-180 kJ


Option: 4

-169.5 kJ


Answers (1)

best_answer

The given \mathrm{\Delta H^{\circ}} value is for the balanced chemical equation, which involves 2 moles of NO reacting.
We need to adjust this \mathrm{\Delta H^{\circ}} value to account for the given reaction when 3 moles of NO react.
Step 1: Identify the given\mathrm{\Delta H^{\circ}} value and the stoichiometric coefficient for NO in the balanced equation:
\mathrm{\Delta H^{\circ}} = −113 kJ/mol (for the balanced equation involving 2 moles of NO)

Step 2: Adjust the\mathrm{\Delta H^{\circ}} value for the reaction involving 3 moles of NO:
Since the reaction involves 1.5 times the moles of NO compared to the balanced equation, we can adjust the \mathrm{\Delta H^{\circ}} value proportionally:

\mathrm{\begin{gathered} \Delta H^{\circ}(\text { adjusted })=\Delta H^{\circ}(\text { original }) \cdot \frac{\text { moles of NO in the given reaction }}{\text { moles of NO in the balanced equation }} \\ \Delta H^{\circ}(\text { adjusted })=-113 \mathrm{~kJ} / \mathrm{mol} \cdot \frac{3 \text { moles }}{2 \text { moles }} \end{gathered}}

Step 3: Perform the calculation:

\mathrm{\Delta H^{\circ}(\text { adjusted })=-169.5 \mathrm{~kJ}}

So, the enthalpy change for the reaction when
3 moles of NO(g) react with excess

\mathrm{\mathrm{O}_2(\mathrm{~g}) \text { toform } 2 \text { molesof } \mathrm{NO}_2(\mathrm{~g})}
is approximately −169.5 kJ.
So, option D is correct

Posted by

Suraj Bhandari

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