Write equilibrium constant expressions for these reactions. For gases, use eithe

Question

Write equilibrium constant expressions for these reactions. For gases, use either pressures or concentrations. a. 3O_2(g) 2O_3(g) b. Fe(s) + 5CO(g) Fe(CO)_3(g) c. Ag_2SO_4(S) 2Ag^+ (aq) + SO_4^2 (aq) Write the expression for K_t for the following reaction. CO_2(g) + H_2(g) CO(g) + H_2O(g) A mixture CO and Cl_2 is placed in a reaction flask that is then sealed. Initially COl = .0102 mol/L and [Cl-2] =.00609 mol/L. When the reaction below CO(g) + Cl_2(g) reaches equilibrium at 600degree K, (Cl_2) = .00301 mol/L. a). Calculate the concentrations of CO and COCl_2 at equilibrium. b). Calculate K_t An equilibrium mixture contains 1.50 mol of nitrogen, 4.45 mol of hydrogen and 7.05 mol of ammonia in a 5.00 L reaction vessel. Calculate K_t, for the Haber-Bosh process. N_2(g) + 3 H_2(g) 2 NH_3(g) H_2(g) and I_2 (g) were added to a heated container, (H_2)_initial = 0.0100 mol/L and [I_2]_initial = 0.00800 mol/L. At equilibrium [I_2] = 0.00560 mol/L. Determine K_e. H_2(g0 + I_2(g) 2 HI(g)

Explanation / Answer

5)

a) 3 02 ---> 2 03

Kc = [02]^2 / [02]^3

b) Fe (s) + 5 CO (g) --- Fe (CO)5 (g)

solids are not considered

so

Kc = [Fe(CO)5] / [C0]^5

c) Ag2S04 (s) ---> 2 Ag+ + S042-

Kc = [Ag+]^2 [S042-]

6)

C02 + H2 --> CO + H20

the equilibrium constant expression is given by

Kc = [CO] [H20] / [C02] [H2]

7)

CO + Cl2 ---> COCl2

using ICE table

initial conc of CO , Cl2 , COCl2 are 0.0102 , 0.00609 , 0

change in conc of CO , Cl2 , COCl2 are -x , -x , +x

equilibrium conc of CO , Cl2 , COCl2 are 0.0102 -x , 0.00609-x , x

given

[Cl2] = 0.00301

so

0.00609 - x = 0.00301

x = 0.00308

so

[CO] = 0.0102 -x = 0.0102 - 0.00308 = 0.00712

[COCl2] = x = 0.00308

now

Kc = [COCL2] / [CO] [Cl2]

Kc = [0.00308] / [0.00712] [0.00301]

Kc = 143.716

so

the value of Kc is 143.716

8)

we know that

concentration = moles / volume (L)

so

at equilibrium

[N2] = 1.5 / 5 = 0.3

[H2] = 4.45 / 5 = 0.89

[NH3] = 7.05 / 5 = 1.41

now

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

Kc = [1.41]^2 / [0.3] [0.89]^3

Kc = 9.4

so

the value of Kc is 9.4

9)

H2 + I2 --> 2HI

using ICE table

initial conc of H2 , I2 , HI are 0.01 , 0.008 , 0

change in conc of H2 , I2 , HI are -x , -x , +2x

equilibrium conc of H2 , I2 , HI are 0.01-x ,0.008-x , 2x

given

[I2] = 0.0056

so

0.008-x = 0.0056

x = 0.0024

now

[H2] = 0.01 - x = 0.01 - 0.0024 = 0.0076

[HI] = 2x = 2 * 0.0024 = 0.0048

now

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

Kc = [0.0048]^2 / [0.0076] [0.0056]

Kc = 0.54

so

the value of Kc is 0.54

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