Question

When a sample of I2(g) (0.07249 mol/L) is placed in 130.0 L reaction vessel at 865.0 K and allowed to come to equilibrium the mixture contains 0.1110 mol/L of I(g). What concentration (mol/L) of I2(g) reacted?
I2(g) = 2I(g)

Answer 1

See your first post below.

Answer 2

The problem is I still don't understand the concept and what is being asked... Anything you can recommend DrBob?? I would appreciate it as I would like to catch up....

Ok so
.............I2(g) ==> 2I
I............ .07249 .......0
C............-x..............2x
E......... .07249-x.....2x


x= 0.1110 mols I
so do I do .07249 - 0.1110 ? But that gives me a negative number...

Answer 3

The problem is I still don't understand the concept and what is being asked... Anything you can recommend DrBob?? I would appreciate it as I would like to catch up....

Ok so
.............I2(g) ==> 2I
I............ .07249 .......0
C............-x..............2x
E......... .07249-x.....2x
<b>Down to here is just great! You have it down perfectly. You messed up on the next step. The problem tells you that mols I at equilibrium = 0.1110 and you've let that be 2x. Therefore,
2x = 0.1110 so x = 0.1110/2 = 0.0555
Then mols I2 = 0.07249-0.0555 = ?(I would round that answer to 0.170 mols) and
M I2 = mols/130.0 L = ?</B>


x= 0.1110 mols I
so do I do .07249 - 0.1110 ? But that gives me a negative number..

Answer 4

Then mols I2 = 0.07249-0.0555 = 0.0170 mols correct?

therefore, m I2 = 0.0170/130= 1.308 x 10^-4 ??

Answer 5

yes