Consider the reaction:
2N2O (g) 2N2 (g) + O2 (g)
The values of Kc at different temperatures are:
Temperature: 838, Kc= 1.10 x 10^-3
Temperature: 1001, Kc= 3.80 x 10^-1
Temperature: 1030, Kc= 8.71 x 10^-1
Temperature: 1053, Kc= 1.67
Which statement is correct at a higher temperature?
A. The forward reaction is favoured.
B. The reverse reaction is favoured.
C. The rate of the reverse reaction is greater than the rate of the forward reaction.
D. The concentration of both reactants and products increase.
Well, well, well, looks like we have a question about chemical reactions and temperature. Hold on tight, because I'm going to bring out the funny chemistry jokes for this one!
Okay, let's analyze the situation. We have a reversible reaction going on here, where N2O is being converted into N2 and O2. The values of Kc at different temperatures are given.
Now, at higher temperatures, things get a little hot-headed, just like me when I'm trying to come up with a good punchline!
If the value of Kc increases as the temperature rises, it means the forward reaction is favored. So, option A seems to be the correct answer in this case.
But hey, don't worry too much if you didn't laugh at my jokes. Chemistry can be a tough nut to crack, just like making people laugh! Keep studying, and I'm sure you'll do great.
Note that as T increases the Kc value increases which means the products are increasing at the expense of the reactants; therefore, A is the only statement that is true.
To determine which statement is correct at a higher temperature, we need to compare the values of the equilibrium constant (Kc) at the different temperatures given in the question.
At a higher temperature, if the value of Kc increases, it indicates that the forward reaction (2N2O → 2N2 + O2) is favored. Conversely, if the value of Kc decreases, it indicates that the reverse reaction (2N2 + O2 → 2N2O) is favored.
Comparing the values of Kc at the given temperatures:
- At 838 K, Kc = 1.10 x 10^-3
- At 1001 K, Kc = 3.80 x 10^-1
- At 1030 K, Kc = 8.71 x 10^-1
- At 1053 K, Kc = 1.67
As the temperature increases from 838 K to 1053 K, the value of Kc continuously increases. Therefore, at a higher temperature, in this case, the correct statement is:
A. The forward reaction is favored.
To determine which statement is correct at a higher temperature, we need to consider the effect of temperature on the equilibrium position of the reaction.
In accordance with Le Chatelier's principle, when the temperature is increased, the equilibrium will shift in the direction that absorbs heat (endothermic direction) to counteract the increase in temperature.
Looking at the given reaction:
2N2O (g) ↔ 2N2 (g) + O2 (g)
We can see that the forward reaction involves the formation of products (N2 and O2) and is, therefore, endothermic. The reverse reaction involves the formation of reactants (N2O) and is exothermic.
Now, let's examine the given values of Kc at different temperatures:
- At temperature 838, Kc= 1.10 x 10^-3
- At temperature 1001, Kc= 3.80 x 10^-1
- At temperature 1030, Kc= 8.71 x 10^-1
- At temperature 1053, Kc= 1.67
Since Kc is the equilibrium constant, its value indicates the relative concentrations of products and reactants at equilibrium. Recall that Kc can be calculated using:
Kc = [products] / [reactants]
Based on the principle that Kc increases as the equilibrium position shifts towards the products, we can deduce the following:
- If Kc > 1, the concentration of products is higher than that of reactants at equilibrium. In this case, the forward reaction is favored.
- If Kc < 1, the concentration of reactants is higher than that of products at equilibrium. In this case, the reverse reaction is favored.
- If Kc = 1, the concentrations of products and reactants are approximately equal at equilibrium.
Now, let's analyze the given values of Kc:
- At temperature 838, Kc = 1.10 x 10^-3 (very small value): This suggests that the concentration of reactants is higher than that of products, indicating that the reverse reaction is favored at this temperature.
- At temperature 1001, Kc = 3.80 x 10^-1 (small value): Again, the concentration of reactants is still higher than that of products, which means the reverse reaction is still favored at this temperature.
- At temperature 1030, Kc = 8.71 x 10^-1 (small value): The concentration of reactants is still higher than that of products, tilting the equilibrium towards the reverse reaction.
- At temperature 1053, Kc = 1.67 (value greater than 1): Now, the concentration of products is higher than that of reactants, indicating that the forward reaction is favored at this higher temperature.
Therefore, the correct statement at a higher temperature is: A. The forward reaction is favored.