Anonymous
Anonymous asked in Science & MathematicsChemistry · 6 months ago

Help me find the percentage of the volume of nitrogen gas N2 (g) and hydrogen gas H2(g)?

The reaction for the Haber process, the industrial production of ammonia, is:

N2(g) + 3H2(g) 🡪 2NH3 (g) When the equilibrium arrived, the NH3(g) has a 25% volume of the Haber. Find the percentage of the volume of nitrogen gas N2 (g) and hydrogen gas H2(g), respectively.

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  • 6 months ago
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    The balanced equation:

     N2(g) + 3H2(g)  → 2NH3(g)

    1 mol N2 reacts with 3 mol H2 to produce 2 mol NH3

     For gases , moles = volumes

    1 volume N2 reacts with 3 volumes H2 to produce 2 volume NH3

     Now you ask for the % volumes when equilibrium has been achieved:

     Let us say that you have Haber reactor that has a volume of 100 L

     You load the reactor with 25 L of N2 and 75 L of H2

    In going to completion :

     25 L N2 reacts with 75 L H2 to produce 50 L NH3

     The reaction proceeds until an equilibrium is achieved , when the NH3 is 25% of the volume of the reactor.

    This means that 25 L  of NH3 has been produced.

    .

    We know that 25 L N2 produces  50 L  NH3 at completion

    Therefore 12.5 L N2 produced 25 L NH3  at equilibrium

    and 37.5 L H2 produce the 25 L NH3

     Therefore , remaining unreacted :

    25L - 12.5 L N2 = 12.5 L N2

    75 L - 37.5 L H2 =  37.5 L H2

    % composition of the gases in the reactor at equilibrium:

    NH3 = 25 L = 25%

    N2 = 12.5 L = 12.5%

     H2 = 37.5 L = 37.5%

  • 6 months ago

    At a constant final pressure in the reaction vessel, the volume % is equal to the mole %.  Since the NH3 is occupying 25%, the N2 and the H2 are occupying 75%, with 1 mole of N2 and 3 moles of H2 for a total of 4 moles.  

    For the N2:

    1/4 X 75% = 18.75%

    For the H2:

    3/4 X 75% = 56.25%

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