Question

#4 - chemical systems and equilibrium (part 1)
an open bottle of pop
question 3 (1 point)
a full syringe contains the equilibrium below. if nitrogen dioxide is brown and the other gas colourless, what is observed when the syringe is used to compress this gas mixture and therefore reduce the volume, increasing the pressure?
2 no₂ (g) <--> n₂o₄ (g)
nothing will happen
the mixture gets darker
the mixture gets lighter
the mixture gets darker and then lighter

Explanation:

Step1: Apply Le Chatelier's Principle

When pressure increases (volume decreases) for the equilibrium \( \ce{2 NO2 (g) <=> N2O4 (g)} \), the system shifts to the side with fewer moles of gas. The left side has 2 moles, the right has 1 mole, so the shift is to the right. But first, compressing the gas increases the concentration of \( \ce{NO2} \) (brown) temporarily, making the mixture darker. Then, as the equilibrium shifts right (consuming \( \ce{NO2} \) to form \( \ce{N2O4} \)), the color lightens, but the initial compression - induced concentration increase makes it darker first, then the equilibrium shift makes it lighter? Wait, no: Wait, when you compress, the immediate effect is higher concentration (so darker), then the equilibrium shifts to reduce pressure (fewer moles), so \( \ce{NO2} \) is consumed, so the mixture gets lighter? But the options: Let's re - evaluate. Wait, the process: When you compress the syringe, the volume decreases. The concentration of all gases increases initially (so \( \ce{NO2} \) concentration goes up, making it darker). Then, according to Le Chatelier's principle, the equilibrium \( \ce{2 NO2 (g) <=> N2O4 (g)} \) (which has 2 moles on left, 1 on right) will shift to the right (fewer moles) to counteract the pressure increase. So \( \ce{NO2} \) (brown) is converted to \( \ce{N2O4} \) (colorless). But the initial compression makes the concentration of \( \ce{NO2} \) higher (darker), then the equilibrium shift reduces \( \ce{NO2} \) concentration (lighter). But does the final color get back to original? No, because the total moles are less, but the initial step is darker, then lighter? Wait, the options: "the mixture gets darker and then lighter" is an option. Wait, let's check the moles: For the reaction \( \ce{2 NO2 (g) <=> N2O4 (g)} \), \( \Delta n = 1 - 2=- 1 \). So increasing pressure (decreasing volume) shifts equilibrium to the right. But the immediate effect of decreasing volume is that the concentration of \( \ce{NO2} \) (and \( \ce{N2O4} \)) increases. So the mixture becomes darker (because \( \ce{NO2} \) is brown) initially. Then, as the equilibrium shifts to the right (consuming \( \ce{NO2} \) to form \( \ce{N2O4} \)), the concentration of \( \ce{NO2} \) decreases, so the mixture becomes lighter (since \( \ce{N2O4} \) is colorless). So the observation is the mixture gets darker (due to increased concentration) and then lighter (due to equilibrium shift).

Step2: Evaluate the Options

• "nothing will happen": Incorrect, as pressure change affects equilibrium.
• "the mixture gets darker": Only the initial step, not the full process.
• "the mixture gets lighter": Incorrect, initial compression makes it darker.
• "the mixture gets darker and then lighter": Correct, as explained: initial concentration increase (darker) then equilibrium shift (lighter).

Answer:

the mixture gets darker and then lighter