Hi guys! I was wondering if someone could help me with this question.
I can immediately eliminate B as an answer since there is no chemical B at the beginning of the reaction. I think the answer isn't A either, since the the concentrations of A and B would probably be the same once equilibrium is reached, but please correct me if I'm wrong. I can't really go much farther than that. I thought about how the reaction quotient is initially 0, since the concentration of B is initially 0 M, but then raises to some positive number as the system reaches equilibrium, but I'm not sure if that is helpful. I suppose there is something important surrounding how the concentration of B eventually becomes greater than [A], but I can't wrap my head around how that would affect the chemical equation i.e. whether the coefficient of A or B would be greater as a result.
Thanks in advance
This is an interesting one, and I'm not convinced we have enough information to answer this - but I'm going to just run through my thinking ANYWAY, and maybe this will help us glean something.
So my first impulse is the answer is A - if it were C and D, I would expect the square factor to make the chemicals A or B to be much smaller. But, we don't know what the equilibrium constant is, so that logic goes out the window.
I then thought - why not make an ICE table? Couldn't hurt. If the equation is A->B:
A B
I 1 0
C -x +x
E 1-x +x
Okay, so [A]~0.2 M, and conc(B)~0.4 M. If I have the stoichiometry correct, then setting these two equal to E SHOULD give the same x:
A: 0.2=1-x <==> x=0.8
B: 0.4=x <==> x=0.4
Okay, so this would suggest the reaction is NOT A->B. What about 2A->B? Well, the ICE table is:
A B
I 1 0
C -2x +x
E 1-2x +x
Which would give:
A: 1-2x=0.2 <==> 2x=0.8 <==> x=0.4
B: x=0.4
Which are the same! So this would suggest to me that the answer is C, 2A-->B.
So yeah, with these kinds of questions, it's worth just playing around and trying random things - I didn't know to use an ICE table, I just decided to make one, and noticed that there was a logical progression I could follow.