1/ calculate the equilibrium constant, Kc, for the following reaction at 25 celc

Question

1/ calculate the equilibrium constant, Kc, for the following reaction at 25 celcius, if the equilibrium concentrations are [H2] = 0.106 M, [I2] = 0.022 M and [HI] = 1.29 M.

H2(g) + I2(g) 2HI(g)

2/ Calculate the equilibrum constant Kc for the following reaction at 25 celcius,

½ HI(g) ¼ H2(g) + ¼ I2(g)

If the equilibrium concentrations are [H2] = 0.106 M, [I2] = 0.022 M and [HI] = 1.29 M

Write the Ka expression for the reaction

3/ At 2000 K , nitrogen gas and oxygen gas combine to form nitric oxide according to the following reaction: N2(g) + O2(g) 2NO(g) H = +1.81 KJ

with an equilibrium constant K of 4.1 x 10^-4. If 0.50 mole of N2 and 0.86 mole of O2 are put into a 2.0 L container at 2000 K , what would the equilibrium concentrations of all species be ?

4/ For the reaction in question #3. How would the following changes effect the equilibrium ? Each change occurs independently of the others. Increase temperature. Decrease pressure. Decrease [O2(g)]. Increase [NO(g)]

5/ The pOH of a solution is 9.75. Calculate the pH, the [H+] and the [OH-] of the solution.

6/ Identify the acid, base, conjugate acid, and conjugate base in the reaction below.

HCHO2(aq) + H2O(I) H3O+(aq) + CHO2-(aq)

8/ Determine the pH and the concentration of all species present in a 1.5 molar solution of hypochlorous acid. The Ka for HCLO is 3.5 x 10^-8.

Explanation / Answer

1. H2(g) + I2(g) 2HI(g)

[H2] = 0.106 M, [I2] = 0.022 M and [HI] = 1.29 M

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

= (1.29)2 / 0.106*0.022

= 713.59

2. ½ HI(g) ¼ H2(g) + ¼ I2(g)

[H2] = 0.106 M, [I2] = 0.022 M and [HI] = 1.29 M

Kc = [H2]1/4 [I2]1/4 / [HI]1/2

= (0.106)1/4 (0.022)1/4 / (1.29)1/2

= 0.570 * 0.385 / 1.135

= 0.1933

Ka expression for the reaction

Ka =  [H2]1/4 [I2]1/4 / [HI]1/2

3.   N2(g) + O2(g) 2NO(g)   H = +1.81 KJ

initially 0.50 0.86 0

at equilibrium 0.50 - 2x 0.86 - 2x 2x

Keq = 4.1 x 10-4

Keq = [NO]2 /.[N2][O2]

4.1 x 10-4 = (2x)2 / (0.50 - 2x)(0.86 - 2x)

4.1 x 10-4 = 4*x2 / (0.43 - 2.72x + 4*x2)

1.763x10-4 - 1.1152x10-3 * x + 1.64x10-3 * x2 =  4*x2

3.998 x2  + 1.1152x10-3 * x - 1.763x10-4 = 0

x = 6.50x10-3

So,

[NO] = 2*6.50x10-3 = 0.013 M

[N2] = 0.50 - 0.013 = 0.487 M

[O2] = 0.86 - 0.013 = 0.847 M

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