Question

3. An industrial chemist introduces 2.0 atm H2 and 2.0 atm CO2 into a 1.00 L container at 25 ⁰C and then increases the temperature to 700 ⁰C. At that temperature, Kp = 0.534. How many grams of H2 are present when equilibrium is established? H2 (g) + CO2 (g) ⇌ H2O (g) + CO (g)

Answer 1

Step-by-step explanation:

H₂ (g) + CO₂ (g) ⇌ H₂O (g) + CO (g)

2 atm pressure at 25° or 298 K . If temperature increases to 700°C or 973K

pressure = 2 x 973 / 298 = 6.53 atm

H₂ (g) + CO₂ (g) ⇌ H₂O (g) + CO (g)

6.53 - x 6.53 - x x x

x² / ( 6.53 - x )² = .534

x / ( 6.53 - x ) = .73

x = 4.77 - .73 x

x = 2.76

pressure of hydrogen gas in equilibrium

= 6.53 - 2.76

= 3.77 atm

moles of hydrogen at this temperature and pressure

n = PV / RT

3.77 x 1 / .082 x 973

= .04725 mole

= .04725 x 2 gram

= .0945 gram

= 94.5 mili gram