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J. W. Hill, R. H. Petrucci, T. W. McCreary, & S. S. Perry General Chemistry, 4th edition.

Problem 14.60.

The equilibrium constant for the isomerization of butane at 25 oC is Kc = 7.94.


If 5.00 g butane is introduced into a 12.5-L flask at 25 oC, what mass of isobutane will be present when equilibrium is reached?




Answer:

Convert the mass of butane (abbreviated B) to molarity, then set up a table of concentrations to solve for the equilibrium concentration of isobutane (abbreviated i-B). Then convert this molarity back to a mass of isobutane.

B: C4H10 molar mass = 4(12.01) + 10(1.01) = 58.14 g/mol


B(g)
i-B(g)
Initial
6.88×10–3
0
Change
–x
+x
Equilibrium
6.88×10–3 – x
x

[i-B]e = 6.11×10–3 M

i-B: C4H10 so the molar mass is the same as butane (which must be true if they are isomers) = 58.14 g/mol.

So the mass of isobutane at equilibrium is:

mass = Molarity×Volume×molar mass = 6.11×10–3 mol/L × 12.5 L × 58.14 g/mol = 4.44 g