The Keq for the equilibrium below is 5.4 × 1013 at 480.0 °C.
2NO (g) + O2 (g) ↔ 2NO2 (g)
What is the value of Keq at this temperature for the following reaction?
NO2 (g) ↔ NO (g) + 1/2 O2 (g)
The Keq for the equilibrium below is 5.4 × 1013 at 480.0 °C.
2NO (g) + O2 (g) ↔ 2NO2 (g)
What is the value of Keq at this temperature for the following reaction?
NO2 (g) ↔ NO (g) + 1/2 O2 (g)
a. 5.4 × 1013
b. 5.66 × 10-3
c. 5.4 × 10-13
d. 1.4 × 10-7
e. none of the above
Well, at least these questions are keeping you on your toes! The value of Keq for the second reaction is related to the value of Keq for the first reaction.
In the first reaction, we have 2 moles of NO combining with 1 mole of O2 to form 2 moles of NO2. So the expression for Keq is (NO2^2) / (NO^2 * O2). Now, in the second reaction, we have 1 mole of NO and 1/2 mole of O2 combining to form 1 mole of NO2. So the expression for Keq in this case is (NO * O2^(1/2)) / NO2.
Now, we need to find the value of Keq for the second reaction at the same temperature as the first reaction. Since the reactions are related by stoichiometry, the value of Keq for the second reaction is simply the inverse of the value of Keq for the first reaction.
Therefore, the value of Keq for the second reaction is 1 / (5.4 × 10^13), which is approximately 1.85 × 10^-14. So, the correct answer is none of the above. Keep going and don't lose your energy!
To find the value of Keq for the reaction:
NO2 (g) ↔ NO (g) + 1/2 O2 (g)
We can use the concept of equilibrium constants. Since NO2 is the product in the first reaction and the reactant in the second reaction, the equilibrium constant (Keq) for the second reaction is the reciprocal of the Keq for the first reaction.
So, the value of Keq for the second reaction is 1 / (5.4 × 10^13) = 1.85 × 10^-14.
Therefore, the correct option is c. 5.4 × 10^-13.
To find the value of Keq for the reaction NO2 (g) ↔ NO (g) + 1/2 O2 (g), you can use the relationship between the Keq values of forward and reverse reactions.
The given reaction, 2NO (g) + O2 (g) ↔ 2NO2 (g), has a Keq of 5.4 × 1013 at 480.0 °C.
The reaction NO2 (g) ↔ NO (g) + 1/2 O2 (g) is the reverse reaction of 2NO (g) + O2 (g) ↔ 2NO2 (g).
According to the principle of microscopic reversibility, the reverse reaction has the reciprocal of the Keq value of the forward reaction.
So, the value of Keq for the reverse reaction NO2 (g) ↔ NO (g) + 1/2 O2 (g) is 1/5.4 × 1013.
However, none of the answer choices provided matches this value exactly. Therefore, the correct answer would be "e. none of the above" since the Keq value for the reverse reaction does not match any of the given options.