Question

The standard molar enthalpy of formation of NH3(g) is -45.9 kJ/mol. What is the enthalpy change if 9.51 g N 2(g) and 1.96 g H2(g) react to produce NH

3(8)?

a.-29.8 kJ/mol-rxn

b.-43.7 kJ/mol-rxn

C.-10.3 kJ/mol-rxn

d.-20.7 kJ/mol-rxn

e. -65.6 kJ/mol-rxn

Answer 1

Answer:

a. -29.8 kJ/mol-rxn

Explanation:

For a chemical reaction system the forward and reverse rate are equal. The standard molar enthalpy formation of NH3 is -45.9 kJ/mol. For the enthalpy of NH3 (8) the molar enthalpy is -29.8kJ/mol. The molar mass of N2 = 28.02g/mol. Molar enthalpy of formation is standard amount of substance produced in the formation of a reaction. The molar enthalpy is the change in enthalpy due to reaction per mole.

Answer 2

The enthalpy change in the reaction to produce ammonia is -29.8 kJ/mol. Thus option A is correct.

The reaction can be:

$\rm N_2\;+\;3\;H_2\;\rightarrow\;2\;NH_3$

1 mole of nitrogen reacts with 3 mole of hydrogen to form 1 mole of ammonia.

The moles of nitrogen are :

moles = $\rm \dfrac{weight}{molecular\;weight}$

Moles of Nitrogen = $\rm \dfrac{9.51}{28}$

Moles of nitrogen = 0.33 moles

The moles of Hydrogen are:

moles of hydrogen = $\rm \dfrac{1.96}{1}$

moles of hydrogen = 1.96 moles

Since 1 moles of nitrogen requires 3 moles of hydrogen.

0.33 moles of nitrogen requires 0.99 moles of hydrogen.

Thus, nitrogen is the limiting reactant.

From the equation, 1 mole of nitrogen results in 2 moles of ammonia.

0.33 moles of nitrogen will results in  = 0.33 $\times$ 2 moles of ammonia.

The formation of ammonia = 0.66 moles.

The standard molar enthalpy of formation of ammonia = -45.9 kJ/mol.

Thus for 1 mole ammonia, enthalpy = -45.9 kJ/mol.

For 0.66 moles of ammonia, enthalpy = 0.66 $\times$ -45.9 kJ/mol.

The enthalpy change = -30.98 kJ/mol

The nearest value is -29.8 kJ/mol for the reaction.

Thus the enthalpy change in the reaction to produce ammonia is -29.8 kJ/mol. Thus option A is correct.

For more information about the enthalpy change, refer to the link:

https://brainly.com/question/1301642