N2 (g) + 3H2 (g) → 2NH3 (g)ΔH = -92.6 Calculate ΔH (in kJ) for the decomposition of 30.5 g NH3 (g) into N2 (g) and H2 (g)
For the reverse reaction, DH = +92.6 (what), is it kJ? If so, then
92.6 kJ x (30.5g/34.0g) = ?
To calculate the enthalpy change (ΔH) for the decomposition of NH3 into N2 and H2, you can use the given enthalpy change for the reaction and the stoichiometry of the balanced equation.
Given: ΔH = -92.6 kJ (from the balanced equation)
Step 1: Determine the molar mass of NH3.
The molar mass of NH3 is calculated by summing the atomic masses of its constituent elements.
Molar mass of N = 14.007 g/mol
Molar mass of H = 1.008 g/mol
Molar mass of NH3 = 14.007 + (3 * 1.008) = 17.031 g/mol
Step 2: Calculate the number of moles of NH3.
Number of moles = mass / molar mass
Number of moles of NH3 = 30.5 g / 17.031 g/mol
Step 3: Use the stoichiometric ratio to determine the number of moles of N2 and H2 produced.
From the balanced equation: N2 (g) + 3H2 (g) → 2NH3 (g)
The stoichiometric ratio for the decomposition of NH3 is 1:2:3 (1 mole of N2, 3 moles of H2, and 2 moles of NH3).
So, for every 2 moles of NH3 decomposed, 1 mole of N2 and 3 moles of H2 are produced.
Step 4: Calculate the number of moles of N2 and H2 produced.
Number of moles of N2 = (number of moles of NH3 / 2) * 1
Number of moles of H2 = (number of moles of NH3 / 2) * 3
Step 5: Calculate the enthalpy change for the decomposition of 30.5 g of NH3.
ΔH = (enthalpy change per mole) * (number of moles of NH3)
ΔH = -92.6 kJ/mol * (number of moles of NH3)
Substitute the value of the number of moles of NH3 calculated in Step 2 into the equation to get the final answer.
By following these steps, you can calculate the enthalpy change (ΔH) for the decomposition of 30.5 g of NH3 into N2 and H2.