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The standard enthalpy change \mathrm{\left(\Delta H^{\circ}\right)} for a reaction is \mathrm{-500 \mathrm{~kJ} / \mathrm{mol}}. Calculate the heat (q) released when 2 moles of the reaction take place at constant pressure.

Option: 1

1000 kJ


Option: 2

500 kJ


Option: 3

-500 kJ


Option: 4

-1000 kJ


Answers (1)

best_answer

The heat (q) released during a reaction at constant pressure is equal to the negative of the enthalpy change \mathrm{\left(\Delta H^{\circ}\right)} multiplied by the number of moles (n) of the reaction:

\mathrm{ q=-\Delta H^{\circ} \cdot n=-(-500 \mathrm{~kJ} / \mathrm{mol}) \cdot 2 \mathrm{~mol}=1000 \mathrm{~kJ} }

So, the correct answer is: A) 1000 \mathrm{~kJ}

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