Question

In this reaction, nitrogen gas combines with hydrogen gas to yield ammonia. The enthalpy (ΔH) of this reaction is -92.22 kJ/mol.For this experiment, 17.15 grams of nitrogen gas and 10.95 grams of hydrogen gas are allowed to react in the reaction vessel. The ammonia vapor that is produced is then condensed, liquefied, and collected into a collection vessel.

Write a balanced thermochemical equation with phase labels for the Haber process with the heat energy as part of the equation.

What is the theoretical yield of ammonia (in grams) if 17.15 grams of nitrogen gas and 10.95 grams of hydrogen gas are allowed to react?

Based on your theoretical yield, what is the percent yield of ammonia if only 11.5 grams of ammonia is produced?

How much heat energy (in kJ) will be absorbed or released if 11.5 grams of ammonia is produced? State whether the energy will be absorbed or released.(State whether the energy will be absorbed or released.)(Show dimensional anaylsis)

Please show work to help me understand. Thanks

Answer 1

A) Given = -92.22 KJ/mol for each 2 mole of NH₃ produced

i.e. the -ve sign indicates that the system is loosing energy.

Hence the Haber's process is an exothermic process.

B) Since ​ is given for 2 mole of ammonia, the balanced equation can be written as followed

C) Given mass of N2= 17.15 gms

Molar mass of N2 = 28.02 g/mol

mass of N2 = 17.15 g

we have below equation to be used:

number of mol of N2,

n = mass of N2/molar mass of N2

=(17.15 g)/(28.02 g/mol)

= 0.6121 mol

Molar mass of H2 = 2.016 g/mol

mass of H2 = 10.95 g

we have below equation to be used:

number of mol of H2,

n = mass of H2/molar mass of H2

=(10.95 g)/(2.016 g/mol)

= 5.432 mol

we have the Balanced chemical equation as:

N2 + 3 H2 ---> 2 NH3 +

1 mol of N2 reacts with 3 mol of H2

for 0.6121 mol of N2, 1.8362 mol of H2 is required

But we have 5.4315 mol of H2

so, N2 is limiting reagent

we will use N2 in further calculation

Molar mass of NH3 = 1*MM(N) + 3*MM(H)

= 1*14.01 + 3*1.008

= 17.034 g/mol

From balanced chemical reaction, we see that

when 1 mol of N2 reacts, 2 mol of NH3 is formed

mol of NH3 formed = (2/1)* moles of N2

= (2/1)*0.6121

= 1.224 mol

we have below equation to be used:

mass of NH3 = number of mol * molar mass

= 1.224*17.03

= 20.9 g

Hence yield of ammonia is 20.9 g.

Calculte the percentage yield.

Percentage yield={(actual yield)/(theoritical yield)}*100

=(11.5/20.9)*100

=55.02%

) From the given data for 1 mole NH₃ formed 46.11 KJ heat is released

Hence for 1.224 moles of NH3 heat released = 1.224 x 46.11= 56.43 KJ

hence for the experiment 56.43 KJ heat is released; i.e H = -56.43 KJ.