Ask Question
9 February, 02:00

Enter your answer in the provided box. An industrial chemist introduces 2.0 atm of H2 and 2.0 atm of CO2 into a 1.00-L container at 25.0°C and then raises the temperature to 700°C, at which Kc = 0.534: H2 (g) + CO2 (g) ⇌ H2O (g) + CO (g) How many grams of H2 are present at equilibrium?

+3
Answers (1)
  1. 9 February, 02:12
    0
    Answer: 0.0944 gram of H2

    Explanation:

    Raising the T from 25 C (298 K) to 700 C (973 K) increases the pressure of each gas by:

    2.0 atm x (973 K / 298 K) = 6.53 atm

    Where

    Kc = Kp because the moles of product equals the moles of reactants.

    At equilibriuim, the amounts are

    P (H2) = 6.53 - x

    P (CO2) = 6.53 - x

    P (H2O) = x

    P (CO) = x

    Kc = Kp =.534 = (x) (x) / [ (6.53 - x) (6.53 - x) ]

    Take the square root of each side

    (.534) ^0.5 = x / (6.53 - x)

    x = 0.731 (6.53 - x)

    x = 4.77 - 0.731x

    1.731x = 4.77

    x = 4.77 / 1.731 = 2.76 atm

    P (H2) at equilibriuim = 6.53 - 2.76 = 3.77 atm

    P (CO2) at equilibrium = 6.53 - 2.76 = 3.77 atm

    PV = nRT

    n = PV/RT = [ (3.77 atm) (1.00 L) ] / [ (0.08206 L atm/K mol) (973 K) ] = 0.0472 mol H2

    0.0472 mol H2 x (2.00 g / 1.00 mol) = 0.0944 g
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “Enter your answer in the provided box. An industrial chemist introduces 2.0 atm of H2 and 2.0 atm of CO2 into a 1.00-L container at 25.0°C ...” in 📗 Chemistry if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers