Consider the reaction H2 (g) + F2 (g) <-- --> 2HF (g) .At a certain temperature,

Question

Consider the reaction H2 (g) + F2 (g) <-- --> 2HF (g) .At a certain temperature, 0.50 moles of hydrogen gas, 0.50 moles of fluorine gas and .25 moles of hydrogen fluoride gas are placed in a 5.0 L container. Determine if the gas is at equilibrium or not and predict the direction it will move in order to get to equilibrium. And, for the mixure of the gases, determine the final concentrations of each gas. (use an appropriate table). Then, determine the partial pressure of each gas at equilibrium and STP (use Dalton's Law). Standard Pressure is 1.00 atm.

Explanation / Answer

H2 (g) +        F2 (g) <-- -->      2HF (g)

0.1                 0.1                      0.05

0.1 - x           0.1 - x                   0.05 + 2x

Concentration of H2 = 0.5 / 5 = 0.1 mol/L

Concentration of F2 = 0.5 / 5 = 0.1 mol/L

Concentration of HF = 0.25 / 5 = 0.05 mol/L

Q = [ HF] ^2 / [H2] [F2]

Q = 0.05^2 / ( 0.1 * 0.1)

Q = 0.25

It is not in equilibrium . In order to get to equilibrium the production of HF should be more so the reaction proceeds from left to right.

Final concentration of H2 = F2 = 0.05 mol/L and final concentration of HF = 0.15 mol/L

Partial pressure of H2 = 0.5 / ( 1.25 ) = 0.4 atm

Partial pressure of F2 = 0.5 / 1.25 = 0.4 atm

Partial pressure of HF = 0.2 atm

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