When F 2 (g) (0.06043 mol/L) and 2.357 mol of Cl 2 (g) in a 39.00 L reaction ves

Question

When F2(g) (0.06043 mol/L) and 2.357 mol of Cl2(g) in a 39.00 L reaction vessel at 665.0 K are allowed to come to equilibrium the mixture contains 137.1 grams of ClF(g). What is the concentration (mol/L) of Cl2(g)?

F2(g)+Cl2(g) = 2ClF(g)

I'm not sure how to go about this problem. Can you please write out all the steps and explain how to solve it? thanks

When F2(g) (0.06043 mol/L) and 2.357 mol of Cl2(g) in a 39.00 L reaction vessel at 665.0 K are allowed to come to equilibrium the mixture contains 137.1 grams of ClF(g). What is the concentration (mol/L) of Cl2(g)?

F2(g)+Cl2(g) = 2ClF(g)

I'm not sure how to go about this problem. Can you please write out all the steps and explain how to solve it? thanks

Explanation / Answer

given

2.357 mol of Cl2

we know that

molarity = moles / volume

molarity of Cl2 = 2.357 / 39 = 6.04 x 10-2

given the reaction is

F2 + CL2 ----> CLF

using ICE table

inital conc of F2 , Cl2 , CLF are 0.06043 , 6.04*10-2 , 0

change in conc of F2 , CL2 , CLF are   -x , -x , 2x

equilbirum conc of F2 , CL2 , CLF are 0.06043 -x , 6.04* 10-2 -x , 2x

so

[ClF]eq =2x

but given

137.1 grams of ClF at equilibrium

moles = mass / molar mass

so

moles of CLF = 137.1 / 54.45 = 2.5179

molarity of ClF = 2.5179 / 39 = 6.456 x 10-2

so

[CLF]eq = 6.456 x 10-2

but we got

[ClF]eq = 2x

so

2x = 6.456 * 10-2

x = 3.228 * 10-2

now

[Cl2}eq = 6.04 *10-2 - 3.228 *10-2

[Cl2]eq = 2.815 *10-2

so conc of Cl2 at equilbiurm is 0.02815 mol/L

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