Question

In the late eighteenth century Priestley prepared ammonia by reacting \(\mathrm{HNO}_{3}(g)\) with hydrogen gas. The thermodynamic equation for the reaction is $$ \mathrm{HNO}_{3}(g)+4 \mathrm{H}_{2}(g) \longrightarrow \mathrm{NH}_{3}(g)+3 \mathrm{H}_{2} \mathrm{O}(g) \quad \Delta H=-637 \mathrm{~kJ} $$ (a) Calculate \(\Delta H\) when one mole of hydrogen gas reacts. (b) What is \(\Delta H\) when \(10.00 \mathrm{~g}\) of \(\mathrm{NH}_{3}(g)\) are made to react with an excess of steam to form \(\mathrm{HNO}_{3}\) and \(\mathrm{H}_{2}\) gases?

Short answer

Answer: (a) The change in enthalpy when 1 mole of hydrogen gas reacts is approximately -159.25 kJ. (b) The change in enthalpy when 10.00 g of ammonia gas reacts is approximately -373.91 kJ.

Step-by-step solution

Part (a): Calculate ΔH for 1 mole of hydrogen gas reacting

To determine the ΔH when one mole of hydrogen gas reacts, we first need to look at the balanced reaction equation: $$ \mathrm{HNO}_{3}(g) + 4 \mathrm{H}_{2}(g) \longrightarrow \mathrm{NH}_{3}(g) + 3 \mathrm{H}_{2} \mathrm{O}(g) \qquad \Delta H = -637 \,\text{kJ} $$ Notice that for this reaction, 4 moles of hydrogen gas are used. Since the ΔH value is given for these stoichiometric amounts, we can calculate the ΔH for just 1 mole of hydrogen gas by dividing the provided value by 4: $$ \Delta H_{1\, \ce{H2}} = \frac{-637 \,\text{kJ}}{4} $$

Calculate the ΔH for 1 mole of hydrogen gas

Now plug in the numbers and find the ΔH for 1 mole of hydrogen gas: $$ \Delta H_{1\, \ce{H2}} = \frac{-637 \, \text{kJ}}{4} \approx -159.25 \, \text{kJ} $$ So, the change in enthalpy when one mole of hydrogen gas reacts is approximately -159.25 kJ.

Part (b): Calculate ΔH for 10.00 g of NH3 reacting

To determine the ΔH for 10.00 g of ammonia gas reacting, we first need to convert grams of ammonia to moles: $$ \text{moles of NH}_3 = \frac{\text{grams of NH}_3}{\text{molar mass of NH}_3} $$ The molar mass of ammonia (\(\ce{NH3}\)) is approximately 17.03 g/mol.

Calculate the moles of NH3

Now plug in the values to find the moles of ammonia: $$ \text{moles of NH}_3 = \frac{10.00 \, \text{g}}{17.03 \, \text{g/mol}} \approx 0.587 \, \text{mol} $$ About 0.587 moles of ammonia gas are reacting.

Calculate the ΔH for the reaction

We can now determine the ΔH for the reaction involving this amount of ammonia by multiplying the ΔH value for the reaction by the ratio of moles of ammonia reacting to the stoichiometric amount in the balanced equation: $$ \Delta H_{\text{for 10.00 g NH}_3} = -637 \, \text{kJ} \times \frac{0.587\, \text{mol}}{1\, \text{mol}} $$

Find the ΔH for the reaction involving 10.00 g of NH3

Now, plug in the numbers and solve for the ΔH for the reaction involving 10.00 g of ammonia: $$ \Delta H_{\text{for 10.00 g NH}_3} = -637 \, \text{kJ} \times \frac{0.587\, \text{mol}}{1\, \text{mol}} \approx -373.91 \, \text{kJ} $$ So, the change in enthalpy when 10.00 g of ammonia gas reacts is approximately -373.91 kJ.