General chemistry equilibrium and acids/bases problems?

These are for a study guide, I have the answers but can't figure out how to do them.

1. Calculate K for the following reaction if a container initially filled only with 0.351 M NOBr has a [NOBr] = 0.298 M at equilibrium. 2NOBr <--->2NO + Br2

Answer: 8.4 x 10^-4

2. If a container is filled with SO2CL2 to an initial pressure of 3.58 atm, what will the equilibrium pressure of SO2, Cl2, and SO2Cl2 be given the reaction SO2 + Cl2 <----> SO2Cl2.

Answer: SO2 = 0.308 atm; Cl2 = 0.308 atm; SO2Cl2 = 3.27 atm

For this one, I calculated Kc=1.06 x 10^3 given Kp = 34.5

3. Calculate K for the following reaction: HClO + F- <----> HF + ClO-

Given: Ka of HClO = 3.0 x 10^-8; Ka of HF = 7.2 x 10^-4

Answer: K = 4.2 x 10^-5

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  • david
    Lv 7
    7 months ago
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    1. Calculate K for the following reaction if a container initially filled only with 0.351 M NOBr has a [NOBr] = 0.298 M at equilibrium. Z2NOBr <--->2NO + Br2

    ======

    assume vol = 1L so that moles of each substance = molarity

    0.351 - 0.298 = 0.053 NOBr is used ... use stoichiometry

    same moles NO are produced [NO] = 0.053

    1/2 moles Br2 are produced ... [Br2] = 0.0265

    Keq = [NO]^2[Br2] / [NOBr]^2

    Keq = (0.053)^2(0.0265) / (0.298)^2 = 8.38 x 10^-4 <<< round

    Answer: 8.4 x 10^-4

    2. If a container is filled with SO2CL2 to an initial pressure of 3.58 atm, what will the equilibrium pressure of SO2, Cl2, and SO2Cl2 be given the reaction SO2 + Cl2 <----> SO2Cl2. ....given Kp = 34.5

    p(SOCl2) at equilib = 3.58 - x

    p(SO2) at equilib = x

    p(Cl2) at equilib = x

    P = nRT/V <<<< if V is constnat then P is directly proportional to n

    Kp = p(SO2Cl2) / [p(SO2)(p(Cl2)]

    34.5 = (3.58 - x) / x^2 ...

    x = 0.308 = pSO2 = pCl2

    3.58 - x = 3.272 = pSO2Cl

    3. Calculate K for the following reaction: HClO + F- <----> HF + ClO-

    Given: Ka of HClO = 3.0 x 10^-8; Ka of HF = 7.2 x 10^-4

    HClO <--> H+ + ClO- (1)

    HF <--> H+ + F- <<<< reverse this reaction

    H+ + F- <--> HF (2) ... add (1) and (2)

    HClO + H+ + F- <--> H+ + ClO- + HF <<<< notice H+ cancels leaving HClO + F- <----> HF + ClO- <<<< which is what you need

    HF(reversed) ... Ka = [HF] / [H+][F-]

    HClO ... Ka = [H+][ClO-] / [HClO]

    .... MULTIPLY both

    ... [H+][ClO-][HF] / [H+][F-][HClO] <<<< cancel [H+]

    K = [HF][ClO-] / [HClO][F-] = (3.0 x 10^-8) / (7.2 x 10^-4)

    K = 4.1666 x 10^-5 ... round

    Answer: K = 4.2 x 10^-5

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