Question

please do each screenshot:

Part A The equlibrium constant, K. for this reaction is 53. H (9) +12(9)=2HI(9) If the equilibrium mixture contains 0.011 M I, and 0.034 M HI, what is the molar concentration of H, ? Express your answer to two significant figures and include the appropriate units. THIẢ + + 3 ? [H2) = Value Units Submit Request Answer

Answer 1

1.

H2 (g) + I2 (g) $\rightleftharpoons$ 2 HI (g)

Kc = [HI]^2 / [H2][I2]

53 = (0.034)^2 / [H2](0.011)

53 X 0.011 = 0.001156 / [H2]

[H2] = 0.001156 / ( 53 X 0.011 )

[H2] = 0.00198 M

[H2] = 1.98 X 10^-3 M

2.

N2O4 (g) $\rightleftharpoons$ 2 NO2 (g)

Kc = [NO2]^2 / [N2O4]

4.9 X 10^-3 = (0.049)^2 / [N2O4]

[N2O4] = 0.002401 / ( 4.9 X 10^-3)

[N2O4] = 0.49 M

[N2O4] = 4.90 X 10^-1 M

3.

3 H2 (g) + N2 (g) $\rightleftharpoons$ 2 NH3 (g)

Kc = [NH3]^2 / [H2]^3[N2]

1.7 X 10^2 = [NH3]^2 / (0.21^3 X 0.018)

[NH3]^2 = 0.02834

[NH3] = square root ( 0.02834 )

[NH3] = 0.168 M

[NH3] = 1.68 X 10^-1 M

5.

O2 (g) + 2 HF (g) + 76 kcal $\rightleftharpoons$ OF2 (g) + H2O (g)

i.

Decrease in volume of the container shifts the equilibrium towards less number of gaseous moles i.e right side in the present case.

Hence, Concentration of H2O will increase.

ii.

Decrease in temperature favours endothermic reaction i.e forward reaction in the present case.

Hence, concentration of H2O will increase.

iii.

Adding more O2, one of the reactants, shifts the equilibrium towards right side.

Hence, concentration of H2O will increase.

iv.

Increasing the volume of the container shifts the equilibrium towards more number of gaseous moles i.e left side in the present case.

Hence, concentration of H2O will decrease.

v.

Adding a catalyst does not alter the equilibrium position. It just makes the reaction to reach equilibrium faster.

Hence no change in concentration of H2O.