2.    The equilibrium constant for HI decomposition at 500 °C is 5.8×10-3.

2HI(g)—–> H2(g) + I2(g)

At that temperature reactants and products are gaseous. (There is no change in the total number of moles, so Kc = Kp.)
A sample of 0.43 mol of HI is placed in a 1.00 L vessel which is then heated to 500 °C. When equilibrium is reached, what is the molar concentration of H2? (Give units).
[H2] =

Iodine is sparingly soluble in pure water. However, it does `dissolve’ in solutions containing excess iodide ion because of the following reaction:

 I–(aq) + I2(aq) I3–(aq)  K = 710

For each of the following cases calculate the equilbrium ratio of [I3] to [I2].

a .   5.00×10-2 mol of I2 is added to 1.00 L of 5.00×10-1 M KI solution.

b.    The solution above is diluted to 12.00 L.

At a particular temperature, K = 1.00×102 for the reaction:

 H2(g) + F2(g) 2HF(g)

In an experiment, at this temperature, 1.90 mol of H2 and 1.90 mol of F2 are introduced into a 1.25-L flask and allowed to react. At equilibrium, all species remain in the gas phase.

What is the equilibrium concentration (in mol/L) of H2?

mol/L

 Tries 0/10

What is the equilibrium concentration (in mol/L) of HF?

mol/L

 Tries 0/10

To the mixture above, an additional 9.50×10-1 mol of H2 is added.
What is the new equilbrium concentration (in mol/L) of HF?

mol/L

Is the statement true or false, with respect to the specified reaction in each case?

 H2(g) + F2(g) 2HF(g)

If a vessel, containing an equilbrium mixture of these gases, is pressurized by adding helium gas, then the equilibrium position will shift to the left.

 H2(g) + F2(g) 2HF(g)

If an equilbrium mixture of these gases is released into a vessel of larger volume, then the equilibrium position will not shift.

 Fe3+(aq) + SCN–(aq) FeSCN2+(aq)

If KOH is dissolved in this solution, then the equilibrium position will shift to the left. (Note that Fe(OH)
3 is insoluble.)

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

If hydrogen gas is added to an equilbrium mixture of these gases, then the equilibrium position will not shift.

 H2O(g) + CO(g) H2(g) + CO2(g) (Exothermic)

If the temperature is increased, then the equilibrium constant for this reaction equation will increase.

Given:

 2CO(g) + O2(g) 2CO2(g) K = 3.01×1091

 C(graphite) + 1/2O2(g) CO(g) K = 2.28×1023

What is the equilibrium constant for the following reaction?

 C(graphite) + O2(g) CO2(g)

Given the following equilibrium constants at 427°C:

 O2(g) + 4Na(l) 2Na2O(s) K1 = 2.50×1049
 O2(g) + 2Na(l) 2NaO(g) K2 = 2.50×109
 O2(g) + 2Na(l) Na2O2(s) K3 = 2.00×1028
 O2(g) + Na(l) NaO2(s) K4 = 3.33×1013

What would be the value of the equilibrium constant for each of the following reactions, at 427°C?

 Na2O2(s) + Na(l) NaO(g) + Na2O(s)

 Tries 0/10
 2NaO2(s) + 2Na(l) 2NaO(g) + Na2O2(s)

##### Do you need a similar assignment done for you from scratch? We have qualified writers to help you. We assure you an A+ quality paper that is free from plagiarism. Order now for an Amazing Discount! Use Discount Code "Newclient" for a 15% Discount!NB: We do not resell papers. Upon ordering, we do an original paper exclusively for you. 