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)
See your first post below.
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...
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
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 = ?
x= 0.1110 mols I
so do I do .07249 - 0.1110 ? But that gives me a negative number..
Then mols I2 = 0.07249-0.0555 = 0.0170 mols correct?
therefore, m I2 = 0.0170/130= 1.308 x 10^-4 ??
yes
To determine the concentration of I2(g) that reacted in the given equilibrium reaction, we can use the stoichiometry of the reaction along with equilibrium concentrations.
The balanced equation provided is:
I2(g) = 2I(g)
Let's assign variables to the concentrations as follows:
[I2] = concentration of I2(g)
[I] = concentration of I(g)
According to the given information, the initial concentration of I2(g) is 0.07249 mol/L. Since I2(g) is the only reactant, its initial concentration is also equal to its equilibrium concentration, denoted as [I2].
The equilibrium concentration of I(g) is given as 0.1110 mol/L. Since the stoichiometry of the reaction is 1:2 between I2(g) and I(g), the change in concentration of I2(g) is twice the change in concentration of I(g), which can be represented as follows:
∆[I2] = 2∆[I]
To find the concentration of I2(g) that reacted, we need to determine the change in concentration (∆[I2]). We can calculate this by subtracting the equilibrium concentration of I2(g) (0.07249 mol/L) from its initial concentration (also 0.07249 mol/L):
∆[I2] = [I2]eq - [I2]initial
Since [I2]eq and [I2]initial are the same in this case, ∆[I2] will be zero. Therefore, no I2(g) reacted.
Hence, the concentration of I2(g) that reacted is 0 mol/L.