Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

The synthesis of ammonia \left(\mathrm{NH}_{3}\right) from nitrogen gas \left(\mathrm{N}_{2}\right)  and hydrogen gas \left(\mathrm{H}_{2}\right) is an important industrial process. It is represented by the following balanced equation:

\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NH}_{3}(\mathrm{~g}) \quad \Delta H^{\circ}=-92.4 \mathrm{~kJ} / \mathrm{mol}
a) Calculate the heat absorbed or released when 150.0 \mathrm{~g} of ammonia is synthesized according to the given reaction.

b) Using the data below, calculate the standard heat of formation \left(\Delta H_{f}^{\circ}\right)$ of ammonia $\left(\mathrm{NH}_{3}\right).

\Delta H_{f}^{\circ} \text { of } \mathrm{N}_{2}(\mathrm{~g})=0 \mathrm{~kJ} / \mathrm{mol}
\Delta H_{f}^{\circ} \text { of } \mathrm{H}_{2}(\mathrm{~g})=0 \mathrm{~kJ} / \mathrm{mol}
\Delta H_{f}^{\circ} \text { of } \mathrm{NH}_{3}(\mathrm{~g})=-46.0 \mathrm{~kJ} / \mathrm{mol}

Option: 1

1.60 \mathrm{~kJ} / \mathrm{mol}


Option: 2

17.0 \mathrm{~kJ} / \mathrm{mol}


Option: 3

92.0 \mathrm{~kJ} / \mathrm{mol}


Option: 4

2.10 \mathrm{~kJ} / \mathrm{mol}


Answers (1)

best_answer

a) The given reaction provides the enthalpy change for the synthesis of 2 moles of \mathrm{NH}_{3}.We need to calculate the enthalpy change for the synthesis of 1 mole of \mathrm{NH}_{3}, and then use it to find the enthalpy change for the given mass of ammonia.

Using the given reaction: \mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NH}_{3}(\mathrm{~g})
\mathrm{\Delta H^{\circ}=-92.4 \mathrm{~kJ} / \mathrm{mol})

For the synthesis of 1 mole of \mathrm{NH}_{3}, the enthalpy change is: \Delta H^{\circ}\left(1\right.$ mole of $\left.\mathrm{NH}_{3}\right)=$ $\frac{\Delta H^{\circ}}{2}=-46.2 \mathrm{~kJ} / \mathrm{mol}
Now, to calculate the heat absorbed or released for 150.0 \mathrm{~g} of \mathrm{NH}_{3} , we'll use the molar mass of \mathrm{\mathrm{NH}_{3}} :

Molar mass of \mathrm{NH}_{3}=14.01 \mathrm{~g} / \mathrm{mol}\left(\mathrm{N}_{2}\right)+3 \times 1.01 \mathrm{~g} / \mathrm{mol}\left(H_{2}\right)=17.04 \mathrm{g} / \mathrm{mol}

\mathrm{Number of moles of \mathrm{NH}_3= Mass \overline{\text { Molar mass }=\frac{150.0 \mathrm{~g}}{17.04 \mathrm{~g} / \mathrm{mol}} \approx 8.80 \mathrm{~mol}}}
Heat absorbed or released =  Number of moles \times \Delta H^{\circ}\left(1\right.$ mole of $\left.\mathrm{NH}_3\right)
Heat =8.80 \mathrm{~mol} \times-46.2 \mathrm{~kJ} / \mathrm{mol} \approx-405.6 \mathrm{~kJ}

b) The standard heat of formation \left(\Delta H_{f}^{\circ}\right) of a compound is the enthalpy change when one mole of the compound is formed from its elements in their standard states.

Using the given data: \Delta H_{f}^{\circ}$ of $\mathrm{N}_{2}(\mathrm{~g})=0 \mathrm{~kJ} / \mathrm{mol}
\Delta H_{f}^{\circ}$ of $\mathrm{H}_{2}(\mathrm{~g})=0 \mathrm{~kJ} / \mathrm{mol}
\Delta H_{f}^{\circ}$ of $\mathrm{NH}_{3}(\mathrm{~g})=-46.0 \mathrm{~kJ} / \mathrm{mol}
\Delta H^{\circ}for the given reaction (synthesis of \mathrm{NH}_{3})\mathrm{=\sum \Delta H_{f}^{\circ} (products) -\sum \Delta H_{f}^{\circ} (reactants)}

\Delta H^{\circ}=2 \times \Delta H_{f}^{\circ}\left(\mathrm{NH}_{3}\right)-\left[\Delta H_{f}^{\circ}\left(\mathrm{N}_{2}\right)+3 \times \Delta H_{f}^{\circ}\left(\mathrm{H}_{2}\right)\right]
\Delta H^{\circ}=2 \times(-46.0 \mathrm{~kJ} / \mathrm{mol})-[0 \mathrm{~kJ} / \mathrm{mol}+3 \times 0 \mathrm{~kJ} / \mathrm{mol}]
\Delta H^{\circ}=-92.0 \mathrm{~kJ} / \mathrm{mol}
Therefore, the standard heat of formation of ammonia \left(\mathrm{NH}_{3}\right)is -92.0 \mathrm{~kJ} / \mathrm{mol}.
Therefore, the correct option is (C).article amsmath

 

Posted by

Divya Prakash Singh

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Latest Question