chemistry

Oh my my test is coming up and I have no idea how to do these questions and when I asked my teacher for help he said you should do it yourself.... Thanks!

An intermediate step in the production of battery acid, sulfuric acid, invoves the information of sulfur trioxide gas from sulfur dioxide and oxygen gases. (this reaction is also the usual first step in the production of acid rain).

The Kp for this reaction at 830degreeC is 0.13. In one experiment, 96.9g of sufur dioxide and 34.6g of oxygen were added to a rxn flask. What must the total pressure in the flask be at eqilibrium if the reaction has an 80% yield of sulfur trioxide. Assume the volume is 3.5L.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Consider the rxn; 4A + B <---> 3C. Initially we had 12 mol of A 12 mol of B in a 2.00L flask, At equilibrium, there is 7.5mol of C in the flask.

Find K and % completion?

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asked by Tri
  1. As I see the the first problem.
    2SO2 + O2 ==> 2SO3.
    Convert g SO2 to mols, convert that to mols SO3, take 80%. That is mols SO3 we should have at the end of the reaction for an 80% yield. Then calculate SO2 remaining, O2 used up (at 80%) and O2 remaining. Add mols of each and use PV = nRT to determine total P.
    Check my thinking.

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    posted by DrBob222
  2. For the second problem, I posted once and it came out garbled so I erased it and started over. Use the ICE chart to solve this. I can't make a good one on the board but hopefully you can understand what I've written.
    4A + B ==> 3C

    I = initial concentrations:
    for A: 12.0 mols/2.00 L = 6.00 M
    for B: 12.0 mols/2.00 L = 6.00 M
    for C: o mols/2.00 L = 0 M

    C = change:
    do this last but I will do the equilibrium and explain what goes in these blanks. You must do change to obtain E.

    E = equilibrium:
    The problem states we had 7.5 mols C at equilibrium. 7.5 mols/2.00 L = 3.50M goes here which means the change for C must be +3.75; i.e., 0 + 3.75 = 3.75.
    If we had +3.75 for change of C, then B must be changed by -3.75/3 = -1.25 M and the final E for compound B is 6.00 - 1.25 = 4.75.
    For A: The change for B was -1.25; therefore, the change for A must be 4 x -1.25 = 5.00 which makes E for compound A 6.00 - 5.00 = 1.00.
    Now just plug all of those equilibrium numbers into the expression for K and solve for K.

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    posted by DrBob222
  3. Note I had a typo:
    E = equilibrium:
    The problem states we had 7.5 mols C at equilibrium. 7.5 mols/2.00 L = 3.50M SHOULD BE 3.75 M goes here which means the change for C must be +3.75; i.e., 0 + 3.75 = 3.75.

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    posted by DrBob222
  4. Everything was good until this happened:

    An intermediate step in the production of battery acid, sulfuric acid, invoves the information of sulfur trioxide gas from sulfur dioxide and oxygen gases. (this reaction is also the usual first step in the production of acid rain).

    The Kp for this reaction at 830degreeC is 0.13. In one experiment, 96.9g of sufur dioxide and 34.6g of oxygen were added to a rxn flask. What must the total pressure in the flask be at eqilibrium if the reaction has an 80% yield of sulfur trioxide. Assume the volume is 3.5L.

    As I see the the first problem.
    2SO2 + O2 ==> 2SO3.
    Convert g SO2 to mols, convert that to mols SO3, take 80%. That is mols SO3 we should have at the end of the reaction for an 80% yield. Then calculate SO2 remaining, O2 used up (at 80%) and O2 remaining. Add mols of each and use PV = nRT to determine total P.
    Check my thinking.

    1. 👍 0
    2. 👎 0
    posted by Tri

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