What is the equilibrium concentration of O_2 in the following reaction if the eq

Question

What is the equilibrium concentration of O_2 in the following reaction if the equilibrium concentrations of N_2 and NO are 24 M and 011 M, respectively. N_2(g) + O_2(g) 2NO(g), K_c = 4.1 times 10^-4 I_2(g) + Br_2(g) 2IBr(g), K_c = 280 a) If a flask starts off as 200 M in both reactants, what will the equilibrium concentrations of all compounds be? b) If a flask starts off is 100 M in I_2 and 300 M in Br_2, what will the equilibrium concentrations of all compounds be? C(s) + CO_2(g) 2CO(g) delta H = 119.8 KJ If the reaction is at equilibrium, how will the reaction shift (if at all) if you a) add more CO_2(g) (and explain why) b) remove CO_2(g) (and explain why) c) remove C(s) (and explain why) d) reduce the volume (and explain why) e) add heat (and explain why)

Explanation / Answer

10) Kc = [NO]2 / [N2] [O2]

4.1 X 10^-4 = (0.011)2 / (0.24) [O2]

[O2] = 1.23 M

11)

                        I2(g) + Br(g) --> 2IBr(g)

Initial              0.2         0.2        0

Change         -x             -x          +2x

Equilibrium    0.2-x        0.2-x        2x

Kc = 280 = [IBr]2 / [I2] [Br2] = (2x)2 / (0.2-x)2

taking square root

16.73 = (2x) / (0.2-x)

3.346 - 16.73x = 2x

3.346 = 18.73x

x = 0.179

The equilibrium concentration will be

[I2] = [Br2] = 0.2 - 0.179 M = 0.021 M

[IBr] = 2x = 2 X 0.179 = 0.358 M

b)             I2(g) + Br(g) --> 2IBr(g)

Initial        0.1         0.3        0

Change         -x             -x          +2x

Equilibrium    0.1-x        0.3-x        2x

Kc = [2x]2 / [0.1-x) (0.3-x) = 280

280 = 4x2 / (0.03 - 0.4x + x2)

2.1 - 28x + 70x^2 = x^2

Solving for x

x = 0.099 M

[I2] = 0.1- 0.099 = 0.01 M

[Br2] = 0.3 - 0.099 M = 0.201 M

[IBr] = 2x = 2 X 0.099 = 0.198 M

12) a) if we will add more CO2 (reactant), so it will move the reaction forward direction, as the reaction will try to nullify the effect of addition of reactant

b) If we will reove CO2, it will move the reaction backward as it will decrease the concentration of reactants

c) There will be no effect of removal of C(s) as it is a solid

d) if we reduce the volume, the concentration will increase and the reaction will move in the direction where there is production of more of gaseous moles

So the equilibrium will move in forward direction

e) The enthalpy of reaction is positive so the reaction is endothermic and on addition of heat the reaction will move in forward direction.

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