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

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80.2K

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Chemistry

At a particular temperature the equilibrium constant for thereaction: H2(g) + F2(g) ⇔ 2HF(g) is K = 81.0. A reaction mixture ina 10.00-L flask contains 0.25 moles each of hydrogen and fluorinegases plus 0.37 moles of HF. What will be the concentration of H2when this mixture reaches equilibrium?

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3.8 Ratings (609 Votes)

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
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