Consider the gas-phase equilibrium A ⇌ B. In a series of experiments, different initial
amounts of A and B are mixed together, and the mixture in each case is allowed to come to
equilibrium. Which one of these experiments would yield values for the amounts of A and B
present at equilibrium different from all the other experiments?
a. 3.0 moles A, 4.5 moles B
b. 4.5 moles A, 3.0 moles B
c. 1.5 moles A, 4.5 moles B
d. 7.5 moles A, no B
e. 0.5 moles A, 7.0 moles B
The answer is given as C....but I have absolutely no idea what is going on here....
No one has answered this question yet.
I will do one of these and let you finish.
A........B
4.75....3.75 for experiment a.
If we start with A = 3.0 and B = 4.5, then we change by x. So A must increase to 3.0 + x and B must decrease by 4.5 - x. Set those equal to obtain
3.0+x = 4.5-x, solve for x and I obtained 0.75. Therefore, at equilibrium A will be 3.0 + 0.75 = 3.75 and B will be 4.5 - 0.75 = 3.75
If you go through the series like that, solving for x in each and put the equilibrium values in a table such as the one I started at the top, you will find that the equilibrium concns are 3.75 for A and 3.75 for B EXCEPT for C where they are 3.00 and 3.00. I must be honest and tell you that I've never seen a problem like this one before and I've been at this chemistry business for more than 65 years. Good luck to you.
To determine which experiment would yield values for the amounts of A and B present at equilibrium different from all the other experiments, we need to consider the concept of the equilibrium constant. The equilibrium constant (K) is a measure of the extent of the reaction at equilibrium and can be calculated using the concentrations or partial pressures of the reactants and products.
In the given equilibrium A ⇌ B, if we assume that A is the reactant and B is the product, the equilibrium constant expression would be:
K = [B] / [A]
Now let's analyze each experiment from the given options:
a. 3.0 moles A, 4.5 moles B
Here, the initial concentration of A is greater than B, suggesting that at equilibrium, more B will be produced. Therefore, the equilibrium composition will have a higher concentration of B compared to the initial amounts.
b. 4.5 moles A, 3.0 moles B
Similar to option a, the initial concentration of A is greater than B. Hence, the equilibrium composition will also have a higher concentration of B compared to the initial amounts.
c. 1.5 moles A, 4.5 moles B
In this case, the initial concentration of A is lower than B. As a result, at equilibrium, A will be favored, leading to a different proportion of A and B.
d. 7.5 moles A, no B
Here, there is no B initially, meaning the reaction cannot proceed in the forward direction. Therefore, at equilibrium, there will be no B present.
e. 0.5 moles A, 7.0 moles B
The initial concentration of B is significantly greater than A. As a result, at equilibrium, more A will be produced, leading to a different composition compared to the initial amounts.
Based on the above analysis, option c (1.5 moles A, 4.5 moles B) would yield values for the amounts of A and B present at equilibrium different from all the other experiments.