Consider the reaction HCHO(g) ?? H2(g) + CO(g) 1.0 mol of HCHO, 1.0 mol of H2 an

Question

Consider the reaction
HCHO(g) ?? H2(g) + CO(g)

1.0 mol of HCHO, 1.0 mol of H2 and 1.0 mol of CO exist in equilibrium in a 2.0 L reaction vessel at 600?C.

a) Determine the value of the equilibrium constant Kc for this system.

Which of the following is correct?

1. Kc cannot be determined for gaseous equilibria.

2. Some other value

3. 2.0

4. 1.0

5. 0.50

2.0 moles of HCHO and 1.0 mol of CO are then added to this system.

b) Which of the following statements about this reaction is now true?

Which of hte following is correct?

1. The reaction mixture is not at equilib- rium, but will move toward equilibrium by forming more HCHO.

2. The forward rate of this reaction is the same as the reverse rate at these new concen- trations.

3. The reaction mixture is not at equilib- rium, but will move toward equilibrium by using up more HCHO.

4. The reaction mixture remains at equilib- rium.

5. The reaction mixture is not at equilib- rium, but no further reaction will occur.

Explanation / Answer

To get the equilibrium constant, K, you need to divide the concentration of the Products by the concentration of the Reactants each raised to their respective coefficient in the balanced reaction. <?xml:namespace prefix = o ns = "urn:schemas-microsoft-com:office:office" /?>

For this rxn the concentration of for each product and reactant is 0.5 mol/L, since there is 1.0 mol/2L.

K= P^coefficient / R^coefficient

K= [0.5] / ([0.5] * [0.5])

K= 2.0

Since K is > 1 more product is present than reactant. Therefore, in order to return to equilibrium more product must be formed.

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