Nitrogen dioxide decomposes on heating according to the equation shown.
\(\mathrm{2NO_2(g)\rightleftharpoons2NO(g)+O_2(g)}\)
When \(4\,\mathrm{mol}\) of nitrogen dioxide were put into a \(1\,\mathrm{dm^3}\) container and heated to a constant temperature, the equilibrium mixture contained \(0.8\,\mathrm{mol}\) of oxygen.
What is the value of the equilibrium constant, \(K_c\), at the temperature of the experiment?
Correct Answer: \( \boxed{\mathrm{D}} \)
At equilibrium:
Since the volume is \(1\,\mathrm{dm^3}\), the concentrations are numerically the same as the number of moles.
The equilibrium expression is:
\(\mathrm{K_c=\dfrac{[NO]^2[O_2]}{[NO_2]^2}}\)
Therefore,
\(\mathrm{K_c=\dfrac{1.6^2\times0.8}{2.4^2}}\)
Therefore, the correct answer is (D).
One particle of X reacts with one particle of Y in a single-step reaction to produce two particles of Z.
This reaction is exothermic and reversible.
\(\mathrm{X+Y\rightleftharpoons2Z}\)
Three statements about the forward and reverse reactions are listed.
1 The activation energy of the forward reaction is equal to the activation energy of the reverse reaction.
2 At equilibrium, the frequency of collisions between one particle of X and one particle of Y is equal to the frequency of collisions between two particles of Z.
3 At equilibrium, the frequency of effective collisions between one particle of X and one particle of Y is equal to the frequency of effective collisions between two particles of Z.
Which statements are correct?
Correct Answer: \( \boxed{\mathrm{D}} \)
Since the reaction is exothermic, the activation energies of the forward and reverse reactions are not equal, so statement 1 is false.
At equilibrium, the total collision frequencies of reactants and products are not necessarily equal, so statement 2 is false.
Dynamic equilibrium means the rate of the forward reaction equals the rate of the reverse reaction. Therefore, the frequency of effective collisions is the same in both directions.
Hence, only statement 3 is correct, so the correct answer is (D).
An equilibrium can be represented by the equation shown.
\(\mathrm{P(aq)+Q(aq)\rightleftharpoons2R(aq)+S(aq)}\)
In a certain mixture, of volume \(1.0\,\mathrm{dm^3}\), the equilibrium concentration of Q is \(10\,\mathrm{mol\,dm^{-3}}\).
What will be the new equilibrium concentration of Q if \(5.0\,\mathrm{mol}\) of pure Q is completely dissolved in the mixture?
Correct Answer: \( \boxed{\mathrm{B}} \)
Adding \(5.0\,\mathrm{mol}\) of Q to a \(1.0\,\mathrm{dm^3}\) solution initially increases the concentration of Q from \(10\) to \(15\,\mathrm{mol\,dm^{-3}}\).
The equilibrium shifts to the right to oppose the increase in Q, consuming some of the added Q to produce more R and S.
Therefore, the final equilibrium concentration of Q is less than \(15\,\mathrm{mol\,dm^{-3}}\) but greater than the original \(10\,\mathrm{mol\,dm^{-3}}\).
Therefore, the correct answer is (B).
