# Given the following equation:

N2 (g) + 3H2 (g) → 2NH3 (g)

ΔH = -92.6 kJ

Calculate ΔH (in kJ) for the decomposition of 30.5 g NH3 (g) into N2 (g) and H2 (g)

## You can work this as a ratio/proportion because you know 2 mols NH3 (2*17=34 g NH3) will require +92.6 kJ (I just changed the sign because the question you asked is the reverse of the equation).

(92.6 kJ/34g) = (? kJ/30.5 g) and solve for ? kJ.

I do these this way.
92.6 kJ x (30.5/2*17) = ? kJ.

## To calculate ΔH for the decomposition of NH3 (g) into N2 (g) and H2 (g), we need to use the given balanced equation and the given ΔH value.

The balanced equation is:
N2 (g) + 3H2 (g) → 2NH3 (g)

1. First, calculate the molar mass of NH3 (ammonia). The molar mass of nitrogen (N) is 14.01 g/mol, and the molar mass of hydrogen (H) is 1.01 g/mol.
Molar mass of NH3 = (3 * molar mass of H) + molar mass of N
= (3 * 1.01) + 14.01
= 3.03 + 14.01
= 17.04 g/mol

2. Calculate the number of moles of NH3:
moles = mass / molar mass
moles = 30.5 g / 17.04 g/mol
≈ 1.79 mol

3. Now, use the balanced equation to relate the moles of NH3 to the moles of N2 and H2.
From the balanced equation, we can see that 2 moles of NH3 decompose to form 1 mole of N2 and 3 moles of H2.
Therefore, 1.79 moles of NH3 will form:
- (1.79 mol / 2) moles of N2
- (1.79 mol / 2) * 3 moles of H2

4. Finally, calculate the heat change using the formula:
ΔH = moles × ΔH value from the balanced equation

- For N2: ΔHN2 = (1.79 mol / 2) × ΔH = (1.79 / 2) × -92.6 kJ/mol
= -82.382 kJ

- For H2: ΔHH2 = (1.79 mol / 2) * 3 × ΔH = (1.79 / 2) * 3 × -92.6 kJ/mol
= -370.767 kJ

ΔHtotal = ΔHN2 + ΔHH2
= -82.382 kJ + -370.767 kJ
≈ -453.149 kJ

Therefore, the ΔH for the decomposition of 30.5 g of NH3 into N2 and H2 is approximately -453.149 kJ.