A vessel of volume 22.4 dm3 contains 2.0 mol h2 and 1.0 mol n2 at 273.15 k initially. all the h2 reacted with sufficient n2 to form nh3. calculate the partial pressures and the total pressure of the final mixture.

Question

Chemistry

Lennon Chen

2 October, 21:08

A vessel of volume 22.4 dm3 contains 2.0 mol h2 and 1.0 mol n2 at 273.15 k initially. all the h2 reacted with sufficient n2 to form nh3. calculate the partial pressures and the total pressure of the final mixture.

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Brady Zamora · Answer 1

Volume = 22.4 dm3

n = 2 mol of H2

n = 1 mol of N2

Temperature = 273.15

All H2 reacts

reaction

N2 + 3H2 = 2NH3

1:3 ratio

Calculation:

N2 initial - N2 reacted = Final N2

1 - 2 * (1/3) = 0.3333 mol of N2 left

H2 = 0 left

NH3 formed = 2/3*1 = 2/3 = 0.666

Total mol:

0.3333 + 0.666 = 1 mol

Apply the equation:

PV = nRT

P = nRT/V = 1*0.0082 * (273.15) / (22.4) = 0.0999924 atm

PH2 = 0

PN2 = 1/3*0.0999924 = 0.0333308 atm

PNH3 = 2/3*0.0999924 = 0.0666616 atm

Answer is 0.0666616 atm