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An industrial chemist introduces 6.8 atm H2 and 6.8 atm CO2 into a 1.00-L container at...

An industrial chemist introduces 6.8 atm H2 and 6.8 atm CO2 into a 1.00-L container at 25.0°C and then raises the temperature to 700.0°C, at which Keq = 0.534: H2(g) + CO2(g) ⇔ H2O(g) + CO(g) How many grams of H2 are present after equilibrium is established?

Solutions

Expert Solution

H2(g) + CO2(g) <----------------> H2O(g) + CO(g)

6.8         6.8                                   0             0

6.8-x       6.8-x                               x              x

Keq = PH2O x PCO / PH2 x PCO2

0.534 = x^2 / (6.8 - x)^2

x = 2.87

PH2O = x = 2.87 atm

PH2 = 6.8 - x = 3.93 atm

from PV = n RT

3.93 x 1 = n x 0.0821 x 973

n = 0.0492

mass of H2 = 0.0492 x 2

                    = 0.0984 g

mass of H2 = 0.0984 g

queen_honey_blossom answered 2 years ago
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