Question #105759
A mixture of H2(g), O2(g) and 2 mL of H2O (l) is present in a 0.500 L rigid container at 25 deg C. The number of moles of H2 and the number of moles of O2 are equal. The total pressure is 1146 mmHg and the vapor pressure of pure water is 24 mmHg. For all questions use the ideal gas constant (R) of 62.4 (L x mmHg)/(K x mol)

a. Determine the partial pressure of H2(g) and the partial pressure of O2(g) in the container.
b. If the reaction occurs at 25 deg C, how many grams of each gas, Hs(g) and O2(g) are present?
c. If the gas were moved to a balloon and the pressure remained constant, what volume would the gas occupy at 15 deg C?
d. If the liquid water is removed and the container is cooled, would the pressure in the container increase, decrease or remain the same. Explain
1
Expert's answer
2020-03-18T12:57:14-0400

(a)Total pressure is 1146mmHg1146mmHg

Vapour Pressure of pure water is 24mmHg24 mmHg

Pressure by O2 and H2 is 1122mmHg1122 mmHg

Mole fraction of H2 is 12\frac{1}{2} .

PH2=12×1122=561mmHg.P_{H_2}=\frac{1}{2}×1122=561mmHg.

PO2=12×1122=561mmHgP_{O_2}=\frac{1}{2}×1122=561mmHg .

(b)PV=nRTPV=nRT

561×0.5=n×62.4×298561×0.5=n×62.4×298

n=0.0151molesn=0.0151moles

MO2=0.0151×32=0.483gmM_{O_2}=0.0151×32=0.483gm

MH2=0.0151×2=0.0302gmM_{H_2}=0.0151×2=0.0302gm

(c)PV=nRTPV=nRT

1122×V=0.0151×2×62.4×2881122×V=0.0151×2×62.4×288

V=0.484L

(d)P/T=constantP/T=constant

P=kTP=kT

If temperature is decreased then pressure decreases.


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS