Question

After 1.60 mol of NH3 gas is placed in a 1600-cm3 box at 25°C, the box is heated to 500 K. At this temperature, the ammonia is partially decomposed to N2 and H2, and a pressure measurement gives 4.85 MPa. Find the number of moles of each component present at 500 K.

Answer 1

Answer:

1.3333 mol for ammonia, 0.1334 mol for N₂ and 0.400 mol for H₂

Explanation:

To calculate the number of mole in the resulting mixture

PV = nRT where is gas constant = 8.314 J/ mol.K T = 500 K Volume = 1600 cm³ = 1600 / 1000000 = 0.0016 m³

substitute the values into the equation

4.85 × 10⁶ Pa × 0.0016 m³ = 8.314 × 500 × n

7760 = 4157 n

n = 7760 / 4157 = 1.8667308 mol

equation of the reaction

2 NH₃(g) → N₂(g) + 3 H₂(g)

2 moles of ammonia yields 1 mole of nitrogen, and 3 moles of hydrogen

1 moles of ammonia will yield 0.5 mole of nitrogen and 1.5 moles of hydrogen

since the ammonia did not fully decomposed

y moles of ammonia will yield 0.5y mole of nitrogen and 1.5y moles of hydrogen

the mole of ammonia remaining = 1.60 - y

sum of the mole in the box after reaction = (1.60 - y) + 0.5 y + 1.5 y = 1.6 + y = 1.8667308 mol

y = 1.8667308 - 1.6 = 0.26673 mol

number of ammonia remaining = 1.6 - 0.26673 mol = 1.3333 mol

number of N₂ present = 0.5 × 0.26673 mol = 0.1334 mol

number of H₂ present = 1.5 × 0.26673 mol = 0.400 mol