Dinitrogen tetroxide (N2O4) is a colourless gas - VCE - SSCE Chemistry - Question 6 - 2005 - Paper 1
Question 6
Dinitrogen tetroxide (N2O4) is a colourless gas. It exists in equilibrium with nitrogen dioxide (NO2), a brown gas. The concentration of NO2 in a gas mixture can be ... show full transcript
Worked Solution & Example Answer:Dinitrogen tetroxide (N2O4) is a colourless gas - VCE - SSCE Chemistry - Question 6 - 2005 - Paper 1
Step 1
a. Write the expression for the equilibrium constant for this reaction.
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Answer
The equilibrium constant expression for the given reaction can be written as:
K = rac{[NO_2]^2}{[N_2O_4]}
where [NO2] and [N2O4] denote the molar concentrations of nitrogen dioxide and dinitrogen tetroxide, respectively.
Step 2
b. i. Keeping the volume constant the temperature is then raised to 357°C. The brown colour then becomes more intense. Is the above reaction (1) exothermic or endothermic? Explain your answer.
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Answer
This reaction is endothermic. When the temperature is increased, the equilibrium shifts to the right to produce more NO2, which is evident as the brown colour intensifies. According to Le Châtelier's principle, an increase in temperature will cause the system to favor the endothermic direction, thus increasing the concentration of NO2.
Step 3
b. ii. Keeping the temperature at 35°C the plunger of the syringe is then pushed in so as to halve the volume. Equilibrium is then re-established. Is the brown colour of the mixture more intense or less intense than before the volume was halved?
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When the volume is halved, the concentration of NO2 will increase because the same amount of substance is now in a smaller volume. This effect makes the equilibrium shift towards the side with fewer moles of gas, which is N2O4. However, since there will still be a significant amount of NO2 present at equilibrium, the colour will remain more intense than before the volume was halved.
Step 4
c. Give the numerical value at 25°C of the equilibrium constant of the reaction.
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Answer
The equilibrium constant at 25°C for the reaction:
NO2(g) ⇌ rac{1}{2} N2O4(g)
is equivalent to half the value of K for the formation of product: