Question

2NO(g) + Br2(g) 2NOBr(g)   A reaction mixture was found to contain 4.70×10-2 moles of NO(g), 4.02×10-2 moles of Br2(g) and 9.28×10-2 moles of NOBr(g), in a 1.00 Liter container. Indicate True (T) or False (F) for each of the following: Indicate True (T) or False (F) for each of the following:

1. In order to reach equilibrium NOBr(g) must be produced.

2. In order to reach equilibrium Kc must decrease.

3. In order to reach equilibrium NO must be produced.

4. Qc is less than Kc.

5. The reaction is at equilibrium. No further reaction will occur.

kc is 154 at 298k

Answer 1

The given equilibrium is

Volume of the reaction container = 1.00 L

The amount of reactants and products in the reaction mixture is

Since the volume in 1.00 L, the concentration and number of moles will be numerically equal.

Hence,

For a generic reaction, , the reaction quotient is expressed as

Now, the reaction quotient of our reaction can be expressed and calculated as follows:

It is also given that the equilibrium constant Kc is 154.

Now, with this information, we can determine whether the given statements are true or false.

1. In order to reach equilibrium NOBr(g) must be produced.

Verdict: True (T)

Reason: Note that the value of reaction quotient is smaller than the equilibrium constant. . Hence, there are more reactants than products compared to the equilibrium composition. Hence, to reach equilibrium, some reactants need to convert to product. Since NOBr(g) is a product, it must be produced to reach equilibrium.

2. In order to reach equilibrium K_C must decrease.

Verdict: False (F)

Reason: Kc is a constant at a particular temperature for a reaction at equilibrium. Hence, its value will not change.

3. In order to reach equilibrium NO must be produced.

Verdict: False (F)

Reason: NO is a reactant in the equilibrium equation. Hence, formation of NO will lead to increase in the denominator in the reaction quotient expression. Hence, Qc value will decrease if NO is produced more. Hence, the reaction will go away from equilibrium to the left as Qc is already smaller than Kc.

4. Qc is less that Kc.

Verdict: True (T)

Reason: We have already calculated the value of Qc above which happens to be smaller than Kc.

97 is less than 154.

5. The reaction is at equilibrium. No further reaction will occur.

Verdict: False (F)

Reason: For the reaction to be at equilibrium, the value of Qc must be equal to Kc. But since it is not the case, the statement must be false.