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What quantity of heat (in kJ) will be released if 0.3987 mol of NH₃ are mixed with 0.200 mol of O₂ in the following chemical reaction?

4 NH₃ (g) + O₂ (g) → 2 N₂H₄ (g) + 2 H₂O (g) ∆H° = -286 kJ/mol

1 Answer

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Answer:

Approximately 56.98 kJ of heat will be released in this chemical reaction when 0.3987 moles of NH₃ are mixed with 0.200 mol of O₂.

Step-by-step explanation:

To calculate the quantity of heat released in this chemical reaction, we need to find the moles of N₂H₄ (hydrazine) formed and then use the given enthalpy change (∆H°) to determine the heat released.

From the balanced chemical equation, we can see that the stoichiometric ratio between NH₃ and N₂H₄ is 4:2, meaning that 4 moles of NH₃ produce 2 moles of N₂H₄.

Step 1: Determine moles of N₂H₄ formed

Given that 0.3987 moles of NH₃ are mixed, we need to find how many moles of N₂H₄ are produced:

0.3987 moles NH₃ × (2 moles N₂H₄ / 4 moles NH₃) = 0.3987 * (2 / 4) moles N₂H₄

0.3987 moles NH₃ → 0.19935 moles N₂H₄

Step 2: Calculate the heat released

Now that we know the number of moles of N₂H₄ formed, we can calculate the quantity of heat released using the enthalpy change (∆H°) given in the reaction.

∆H° = -286 kJ/mol (given)

The heat released can be calculated as follows:

Heat released = ∆H° × moles of N₂H₄ formed

Heat released = -286 kJ/mol × 0.19935 moles N₂H₄

Now, perform the calculation:

Heat released ≈ -56.98 kJ

So, approximately 56.98 kJ of heat will be released in this chemical reaction when 0.3987 moles of NH₃ are mixed with 0.200 mol of O₂.

User Mehdi Ijadnazar
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