When F2(g) (0.02028 mol/L) and 0.02028 mol/L of Cl2(g) in a 460.0 L reaction vessel at 774.0 K are allowed to come to equilibrium the mixture contains 0.02898 mol/L of ClF(g). What concentration (mol/L) of F2(g) reacted?
F2(g)+Cl2(g) = 2ClF(g)
3 answers
See your first post below.
.............F2(g) + Cl2(g) ==> 2ClF(g)
I........0.02028 ...... 0.02028 ........ 0
C...........x....................x.............. -2x
E.....0.02028+x ..... 0.02028+x...... -2x
2x = 0.02898 =
x = .01449
therefore 0.02028 - .01449 = .00579?
I........0.02028 ...... 0.02028 ........ 0
C...........x....................x.............. -2x
E.....0.02028+x ..... 0.02028+x...... -2x
2x = 0.02898 =
x = .01449
therefore 0.02028 - .01449 = .00579?
The reaction is going from left to right so it must be 0.02028-x. You MUST be using up x amount and not adding x amount.
Finally, notice that the problem asks for concn F2 REACTED?. The amount reacted is x so 0.01449 is what reacted. The o.00579 is what is left after the x amount has reacted.
Finally, notice that the problem asks for concn F2 REACTED?. The amount reacted is x so 0.01449 is what reacted. The o.00579 is what is left after the x amount has reacted.
9 answers