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)
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?
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.
To find the concentration of F2(g) that reacted, we'll first need to calculate the initial concentrations of F2(g) and Cl2(g). Then, we'll use the equilibrium concentrations to calculate the change in concentrations, and finally, subtract the change from the initial concentration to find the concentration of F2(g) that reacted.
1. Calculate the initial concentrations of F2(g) and Cl2(g):
Initial concentration of F2(g) = 0.02028 mol/L
Initial concentration of Cl2(g) = 0.02028 mol/L
2. Use the given equilibrium equation to determine the stoichiometric relationship between F2(g) and Cl2(g):
The balanced equation shows that 1 mole of F2 reacts with 1 mole of Cl2 to form 2 moles of ClF.
3. Calculate the change in concentrations of F2(g) and Cl2(g):
Since 1 mole of F2 reacts with 1 mole of Cl2, the change in concentration for F2(g) will be -2x (where x is the concentration of F2(g) that reacted), and the change in concentration for Cl2(g) will be -x.
4. Use the equilibrium concentrations of ClF(g) and the stoichiometric relationship to find the change in concentration of Cl2(g):
The equilibrium concentration of ClF(g) is given as 0.02898 mol/L.
Since the stoichiometric relationship shows that 1 mole of F2 reacts with 1 mole of Cl2 to form 2 moles of ClF, the change in concentration of Cl2(g) will be equal to 2 times the change in concentration of ClF(g), which is 2 * x.
5. Write the expression for equilibrium concentrations in terms of initial concentrations and changes:
For F2(g): equilibrium concentration = initial concentration - 2x
For Cl2(g): equilibrium concentration = initial concentration - x
6. Substituting the given values into the expressions for equilibrium concentrations:
For Cl2(g): 0.02028 - x = 0.02898 (equation 1)
7. Solve equation 1 for x:
x = 0.02028 - 0.02898
x = -0.0087 mol/L
8. The negative value for x indicates that there was an excess of Cl2 present, and no F2 reacted. Therefore, the concentration of F2(g) that reacted is 0 mol/L.
In conclusion, the concentration of F2(g) that reacted is 0 mol/L.