H2(g) + I2(g) 2HI(g) A 1.00-L flask contains a mixture of 2.00 mol H2 and 2.00 m
ID: 490441 • Letter: H
Question
H2(g) + I2(g) 2HI(g) A 1.00-L flask contains a mixture of 2.00 mol H2 and 2.00 mol I2 at 698K. If the equilibrium concentration of hydroiodic acid is 3.16M, what is Kc at 698K? H2(g) + I2(g) 2HI(g) A 1.00-L flask contains a mixture of 2.00 mol H2 and 2.00 mol I2 at 698K. If the equilibrium concentration of hydroiodic acid is 3.16M, what is Kc at 698K? H2(g) + I2(g) 2HI(g) A 1.00-L flask contains a mixture of 2.00 mol H2 and 2.00 mol I2 at 698K. If the equilibrium concentration of hydroiodic acid is 3.16M, what is Kc at 698K?Explanation / Answer
H2 + I2 <-------------> 2HI
2.0 2.0 0 initial concentration
2.0-x 2.0-x 2x equilibrium concentration
Given 2x = 3.16 M , hence x = 1.58 M
Thus the equilibrium concentration
0.42 0.42 3.16 equilibrium concentration
Thus
Kc = (3.16)2/0.42 x0.42
= 56.60
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