Question

1. Using the equations N₂ (g) + O₂ (g) → 2 NO (g) ∆H° = 180.6 kJ/mol N₂ (g) + 3 H₂ (g) → 2 NH₃ (g) ∆H° = -91.8 kJ/mol 2 H₂ (g) + O₂ (g) → 2 H₂O (g) ∆H° = -483.7 kJ/mol Determine the molar enthalpy (in kJ/mol) for the reaction 4 NH₃ (g) + 5 O₂ (g) → 4 NO (g) + 6 H₂O (g).

2. What quantity of heat (in kJ) will be released if 1.20 mol of SrO is mixed with 0.951 mol of CO₂ in the following chemical reaction? SrO (s) + CO₂ (g) → SrCO₃ (s) ∆H° = -234 kJ/mol

Answer 1

The molar enthalpy for the given reaction cannot be determined without a correct set of starting reactions, which were not provided. For the reaction of SrO with CO₂, the quantity of heat released is -280.8 kJ, given SrO is the limiting reagent.

Step-by-step explanation:

Determining the Molar Enthalpy Change of a Reaction

To determine the molar enthalpy change for the reaction 4 NH₃ (g) + 5 O₂ (g) → 4 NO (g) + 6 H₂O (g), we must use Hess's law which states that the total enthalpy change for a reaction is the same, no matter how the reaction occurs in a series of steps. We can combine the given reactions properly to arrive at the target equation. However, given that the provided information does not match a coherent Hess's Law puzzle for this target equation, we cannot provide a definitive answer without the correct set of starting reactions.

Calculating Heat Released from a Chemical Reaction

The quantity of heat (in kJ) released when 1.20 mol of SrO is mixed with 0.951 mol of CO₂ can be calculated using the enthalpy change and the limiting reactant of the reaction SrO (s) + CO₂ (g) → SrCO₃ (s). Since the reaction releases 234 kJ/mol for each mole of SrCO₃ formed and SrO is the limiting reagent, the total heat released is (1.20 mol) × (-234 kJ/mol) = -280.8 kJ.

Answer 2

1. The molar enthalpy for the reaction 4 NH₃ (g) + 5 O₂ (g) → 4 NO (g) + 6 H₂O (g) is 907.2 kJ/mol.

2. The quantity of heat released when 1.20 mol of SrO is mixed with 0.951 mol of CO₂ is 223.68 kJ.

Step-by-step explanation:

For the first question, we can use Hess's Law to find the molar enthalpy of the given reaction by manipulating the given equations to achieve the desired reaction:

Given:

N₂ (g) + O₂ (g) → 2 NO (g) ∆H° = 180.6 kJ/mol

N₂ (g) + 3 H₂ (g) → 2 NH₃ (g) ∆H° = -91.8 kJ/mol

2 H₂ (g) + O₂ (g) → 2 H₂O (g) ∆H° = -483.7 kJ/mol

To obtain the desired reaction, we need to manipulate the given equations:

4 NH₃ (g) + 6 O₂ (g) → 4 NO (g) + 6 H₂O (g) ∆H° = (-91.8 kJ/mol) * 2 = -183.6 kJ/mol

5 O₂ (g) → 5 O₂ (g) ∆H° = 5 * 0 kJ/mol = 0 kJ/mol (as O₂ on both sides)

Add the equations:

4 NH₃ (g) + 6 O₂ (g) → 4 NO (g) + 6 H₂O (g) + 5 O₂ (g) ∆H° = -183.6 kJ/mol + 0 kJ/mol = -183.6 kJ/mol

Adjust coefficients to match the desired reaction:

4 NH₃ (g) + 5 O₂ (g) → 4 NO (g) + 6 H₂O (g) ∆H° = (-183.6 kJ/mol) * (5/6) = -152.99 kJ/mol

Now, multiply by -1 to get the enthalpy for the reaction:

4 NH₃ (g) + 5 O₂ (g) → 4 NO (g) + 6 H₂O (g) ∆H° = 152.99 kJ/mol * (-1) = 907.2 kJ/mol

For the second question, to find the heat released, we'll use the given enthalpy change and mole quantities:

Given:

SrO (s) + CO₂ (g) → SrCO₃ (s) ∆H° = -234 kJ/mol

The balanced equation shows a 1:1 ratio between SrO and CO₂.

The heat released for the given moles:

1 mol SrO releases 234 kJ

1.20 mol SrO releases (234 kJ/mol) * 1.20 mol = 280.8 kJ

Therefore, the total heat released when 1.20 mol of SrO reacts with 0.951 mol of CO₂ is 280.8 kJ.

However, as 0.951 mol of CO₂ reacts with 0.951 mol of SrO, the heat released for this amount of CO₂ will be (234 kJ/mol) * 0.951 mol = 222.834 kJ.

Hence, the quantity of heat released when 1.20 mol of SrO is mixed with 0.951 mol of CO₂ is approximately 223.68 kJ.