A malleable is the correct option.
Malleability is a physical property.
Answer:
- Option <u><em>C) The rates of the forward and reverse reactions are equal.</em></u>
Explanation:
NO₂ and N₂O₄ undergo the following <em>equilibrium</em> reaction:
That is a reversible reaction, i.e. there are two simultaneous reactions: the direct or forward reaction and the reverse reaction:
- Direct reaction: 2NO₂(g) → N₂O₄(g)
- Reverse reaction: 2NO₂(g) ← N₂O₄(g)
At the beginning, only NO₂(g) is in the sealed container. The NO₂ concentration is maximum, and the rate of the forward reaction is maximum.
As the reaction progresses, the concentration of NO₂ diminishes, and, consequently, the rate of the forward reaction decreases.
As soon as the N₂O₄ appears, the reverse reaction starts. At the beginning the rate is low, but as the N₂O₄ concentration increases the rate of the reverse reaction increases.
When both forward and reverse rates become equal the equilibrium has been reached. This is what is called a dynamical equilibrium.
Then, as per the choices, you have that, at equilibrium:
<u>A) No N₂O₄ is present</u>:
- False: as explained above, at equilibrium both NO₂ and N₂O₄ are present.
<u>B) No chemical reactions are occurring</u>.
- False: as explained above, at equilibrium both forward and reverse reaction are occurring at the same rate.
<u>C) The rates of the forward and reverse reactions are equal</u>.
- True: as explained, this is the meaning of dynamic equilibrium.
<u>D) The maximum number of molecules has been reached</u>.
- False: the number of molecules of each compound at equilibrium will be given by the constant of equiibrium, Keq = [N₂O₄] / [NO₂]², and this value varies with the temperature.
Answer:
removing the Cl₂ as it is formed
.
adding more ICl(s)
.
removing some of the I₂(s).
Explanation:
<em>Le Châtelier's principle </em><em>states that when there is an dynamic equilibrium, and this equilibrium is disturbed by an external factor, the equilibrium will be shifted in the direction that can cancel the effect of the external factor to reattain the equilibrium.</em>
<em />
<u>1) Decreasing the volume of the container:</u>
- Decreasing the volume of the container will increase the pressure.
- When there is an increase in pressure, the equilibrium will shift towards the side with fewer moles of gas of the reaction. And when there is a decrease in pressure, the equilibrium will shift towards the side with more moles of gas of the reaction.
- The reactants side (left) has no moles of gases and the products side (right) has 1.0 mole of gases.
- So, increasing the pressure will shift the reaction to the side with lower moles of gas (left side) and so the total amount of Cl₂ produced is decreased.
so, decreasing the volume of the container will decrease the total amount of Cl₂ produced.
<u>2) Removing the Cl₂ as it is formed:</u>
- Removing Cl₂ gas will decrease the concentration of the products side, so the reaction will be shifted to the right side to suppress the decrease in the concentration of Cl₂ gas by removing and so the total amount of Cl₂ produced is increased.
so, removing the Cl₂ as it is formed will increase the total amount of Cl₂ produced.
<u><em>3) Adding more ICl(s)
:</em></u>
- Adding ICl(s) will increase the concentration of the reactants side, so the reaction will be shifted to the right side to suppress the increase in the concentration of ICl(s) by addition and so the total amount of Cl₂ produced is increased.
so, adding more ICl(s) will increase the total amount of Cl₂ produced.
<u>2) Removing some of the I₂(s):</u>
- Removing I₂ gas will decrease the concentration of the products side, so the reaction will be shifted to the right side to suppress the decrease in the concentration of Cl₂ gas by removing and so the total amount of Cl₂ produced is increased.
so, removing some of the I₂(s) will increase the total amount of Cl₂ produced.
<em>the following changes will increase the total amount of of Cl2 that can be produced:</em>
- removing the Cl₂ as it is formed
.
- removing some of the I₂(s).
A. sodium and chlorine b. carbon and hydrogen c. nitrogen and hydrogen d. bromine
