What is the standard enthalpy change for the reaction N2(g) + 3H2(g) ⇌ 2NH3(g) i

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Question: What is the standard enthalpy change for the reaction N2(g) + 3H2(g) ⇌ 2NH3(g) if ΔHf°(NH3) = -45.9 kJ/mol?

Options:

Correct Answer: -91.8 kJ

Solution:

ΔH = 2 * ΔHf°(NH3) = 2 * (-45.9) = -91.8 kJ

What is the standard enthalpy change for the reaction N2(g) + 3H2(g) ⇌ 2NH3(g) if ΔHf°(NH3) = -45.9 kJ/mol?

What is the standard enthalpy change for the reaction N2(g) + 3H2(g) ⇌ 2NH3(g) if ΔHf°(NH3) = -45.9 kJ/mol?

Step 1: Identify the reaction: N2(g) + 3H2(g) ⇌ 2NH3(g).
Step 2: Recognize that ΔHf°(NH3) is the standard enthalpy of formation for ammonia (NH3), which is given as -45.9 kJ/mol.
Step 3: Since the reaction produces 2 moles of NH3, we need to calculate the total enthalpy change for the formation of 2 moles.
Step 4: Multiply the ΔHf°(NH3) value by the number of moles produced: 2 * ΔHf°(NH3) = 2 * (-45.9 kJ/mol).
Step 5: Perform the multiplication: 2 * (-45.9) = -91.8 kJ.
Step 6: Conclude that the standard enthalpy change (ΔH) for the reaction is -91.8 kJ.

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