00150 m at equilibriumh2 i2 0000165 m h2 g ci 0m i2

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Unformatted text preview: 2 2 Set (2): Initially: [HI] = 0.00150 M At equilibrium:[H2] = [I2] = 0.000165 M H2 (g) Ci 0M + I2 (g ) 0M [0.00150]2 Q= = infinity [0][0] Cf 0.000165 2 HI ( g ) 0.000165 0.00150 M 0.00150 Q K 0.00150 - 2×0.000165 (0.0015 − 2 × 0.000165) 2 K= = 50.3 2 (0.000165) Same as (1) Calculate K and Q based on the given three separate sets based of experimental data for: of H (g) + I (g) 2 HI ( g ) 2 2 Set (3) Initially: [H2] = [I2] = [HI] = 0.00150 M Set Initially: At equilibrium:[H2] = 0.000496 M 0.000496 H2 (g) Ci 0.00150 M 0.00150 + I2 (g ) 2 HI ( g ) 0.00150 M 0.00150 M 2 [0.0015] Q= =1 [0.0015][0.0015] Cf 0.000496M 0.000496M Q<K 0.00150 + 2×(0.0015-0.000496) (0.00150 + 2 × (0.00150 − 0.000496)) 2 K= = 50.1 2 (0.000496) Same as (1) & (2) Summary for this problem Summary • K values in 3 cases are about the same (around 50) H 2 ( g ) + I2 ( g ) 2 HI ( g ) • Q values are very different (from 0 to infinity). Set (1) starts with Q= 0 (reactants only); ends with Q = K Set (2) starts with Q = ∞ (products only); ends with Q = K Set Set (3) starts in the middle ends with Q = K This experiment proves the law of mass action: A reaction will reach the same point at the end (K), but the starting conditions (Q) may be different....
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This document was uploaded on 02/26/2014.

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