Post a New Question

General Chemistry

posted by .

At a certain temperature, the equilibrium constant for the following chemical equation is 2.90. At this temperature, calculate the number of moles of NO2(g) that must be added to 2.53 mol of SO2(g) in order to form 1.10 mol of SO3(g) at equilibrium.

SO2(g)+ NO2(g) <> SO3(g) + NO(g)


I tried doing the ICE chart but there seems to be something wrong with my calculation. -_-

  • General Chemistry -

    Try this

    ..........SO2 + NO2 ==> SO3 + NO
    I.........2.53...0.......0.....0
    C..........-x....x.......1.1...1.1
    E.........2.53-x..x......1.1....1.1

    Total mols = 2.53-x_x_1,1+1.1 = 4.73-x

    XSO2 = (2.53-x)/(4.73-x)
    XNO2 = x/(4.73-x)
    XSO3 = 1.1/(4.73-x)
    XNO = 1.1(4.73-x)

    Then partial pressures are
    pSO2 = (2.53-x)/(4.73-x)]*Ptotal
    pNO2 = [x/(4.73-x)]*Ptotal
    pSO3 = [1.1/(4.73-x)]*Ptotal
    pNO = [1.1/4.73-x)]*Ptotal

    I won't finish but here is what you do.
    Substitute partial pressures from above into K expression. Solve for x = mols NO2 and you have it. Don't worry about all of this extraneous stuff; it cancels and you are left with a simple quadratic to solve.

Answer This Question

First Name
School Subject
Your Answer

Related Questions

More Related Questions

Post a New Question