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At a particular temperature the equilibrium constant for the reaction: H2(g) + F2(g) ⇔ 2HF(g) is...

At a particular temperature the equilibrium constant for the reaction: H2(g) + F2(g) ⇔ 2HF(g) is K = 81.0. A reaction mixture in a 10.00-L flask contains 0.25 moles each of hydrogen and fluorine gases plus 0.37 moles of HF. What will be the concentration of H2 when this mixture reaches equilibrium?

Solutions

Expert Solution

molarity of H2 = 0.25 / 10 = 0.025 M

molarity of F2 = 0.37 / 10 = 0.037 M

H2    +      F2 <-------------------------> 2HF

0.025   0.037                                  0      ---------------------> initial

0.025-x      0.037-x                             2x   ------------------------> equilibrium

K = (2x)^2 / (0.025-x) (0.037-x)

81.0 = (2x)^2 / (0.025-x) (0.037-x)

81.0 = 4x^2 / (0.025-x) (0.037-x)

x =0.023

H2 concentration at equilibrium = 0.025-x = 0.025- 0.023

H2 concentration at equilibrium     = 0.002 M

queen_honey_blossom answered 7 months ago
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