Question

1. A container is filled with 16.0 g of O₂ and 14.0 g of N₂.

a. What is the volume of the container at STP?

b. What is the partial pressure of the O₂ gas?

c. What is the mole fraction of N₂ in the mixture?

2. Look carefully at the following reaction: N₂(g) + 3H₂(g) -> 2NH₃(g)

How will the enthalpy change (H) compared to the internal energy change (E) for this reaction? That is, will H be equal to, greater than, or less than E? Explain.

Answer 1

mol of O2 = mass/MW = 16/32 = 0.5

mol of N2 = mass/MW = 14/28 = 0.5

Total mol = 0.5+0.5 = 1

At STP. 1 mol occupies 22.4 L of space

therefore

a) STP ratio = 22.4L

b)

P of O2

Px = xO2*PT

Px = 0.5*1atm = 0.5 atm of O2

c)

mole fraction of N2 = mol of N2 / total mol = 0.5/1 = 0.5

2)

N2(g) + 3H2(g) -> 2NH3(g)

H = U + PV

dH = dU + dPdV

from this example, since number of gas mole decrease drastically (from 4 to 2), then expect dV (to be negative)

Then dH = dU - dP*dV

E reaction decreases