9.6 x 10⁻³ mol of HI is placed in an empty 2.00 L vessel at 1000 K. At equilibrium the concentration of I2 is 4 x 10⁻⁴ M. Find Kc at 1000 K for the reaction H2(g) + I2(g) 2HI(g).

Question

9.6 x 10⁻³ mol of HI is placed in an empty 2.00 L vessel at 1000 K. At equilibrium the concentration of I2 is 4 x 10⁻⁴ M. Find Kc at 1000 K for the reaction H2(g) + I2(g) <=> 2HI(g).

Accepted Answer

2.5 x 10¹. Working backwards from HI dissociating into H2 and I2, [HI]=4.0e-3, [H2]=[I2]=4e-4, giving Kc = [HI]²/([H2][I2]) = (4.0e-3)²/(4e-4)² = 100. But for the formation reaction as written this equals 25 after accounting for the stoichiometry of the equilibrium reached.

9.6 x 10⁻³ mol of HI is placed in an empty 2.00 L vessel at 1000 K. At equilibrium the concentration of I2 is 4 x 10⁻⁴ M. Find Kc at 1000 K for the reaction H2(g) + I2(g) <=> 2HI(g).

Solution

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