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lecture+7sf - Equilibrium Assume we mix 1.00 mol H2(g) and...

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Equilibrium Assume we mix 1.00 mol H 2 (g) and 1.00 mol I 2 (g) in a 1.00 L container. The reactants react to form HI(g). H 2 (g) + I 2 (g) → 2HI(g) When the reaction is over, how many moles of HI form? The obvious answer is 2.00 mol HI. This answer assumes that the reaction goes to completion, leaving 0 moles of H 2 and I 2 left over. In a closed system, the reaction will not go to completion. As the reaction proceeds, the rate of the forward reaction (from left to right) decreases as the concentration of reactants decrease (collisions involving H 2 and I 2 become less frequent). As products (in this case HI) build up, the rate of the reverse reaction (from right to left) increases (collisions involving HI molecules become more likely). There reaches a point when the rate of the forward reaction equals the rate of the reverse reaction. This is called equilibrium . How far the reaction goes to reach equilibrium depends on the individual reacting system. Without additional information, in the example above we can only say that at equilibrium, less than 2.00 mol HI will be formed. There will reach some point (between 0 and 2.00 mol HI) where the rates of the forward and reverse reactions are equal. A reaction proceeds until it reaches equilibrium. What would we observe that tells us that equilibrium is reached? When equilibrium is reached, the reaction appears to be over. There would be no further changes in any macroscopic properties. Pressure, temperature, amounts, would all stay the same. On a molecular level, however, the reaction is still going on. However, the rates of the forward and reverse reactions are equal, leaving a net rate of zero. So on the macroscopic level, no further changes occur.
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Quantitative Measure of Equilibrium Let’s look at some data on the reaction between H 2 and I 2 . H 2 (g) + I 2 (g) 2HI(g) Initial Equilibrium Expt [H 2 ] [I 2 ] [HI] React [ H 2 ] [I 2 ] [HI] 1 1.00 1.00 0 0.787 0.213 0.213 1.574 2 0 0 1.00 0.214 0.107 0.107 0.786 3 1.00 1.00 1.00 0.680 0.320 0.320 2.36 In experiment 1, we start with 1.00 M each of H 2 , I 2 and find that 0.787 M react, leaving over 0.213 M each of H 2 , I 2 at equilibrium and forming 0.787(2) or 1.574 M HI, where the factor of 2 is due to the 2:1 mol ratio in the balanced equation. In experiment 2, we start with 1.00 M HI and find that 0.214 M react, leaving over 0.786 M HI at equilibrium and forming 0.214(1/2) or 0.107 M H 2 and I 2 , where the factor of 1/2 is due to the 1:2 mol ratio in the balanced equation. In experiment 3, we start with 1.00 M each of H 2 , I 2 , HI and find that 0.680 M of H 2 and I 2 react, leaving over 0.320 M each of H 2 , I 2 at equilibrium and forming 0.680(2) or 1.36 M HI, where the factor of 2 is due to the 2:1 mol ratio in the balanced equation. The 1.36 M HI formed is added to the original 1.00 M HI present, making 2.36 M HI at equilibrium.
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We have three different equilibrium conditions for the same reaction at the same temperature. There is one expression, called the equilibrium constant, which is the same for all three sets of
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This note was uploaded on 03/08/2011 for the course CHEM 162 taught by Professor Siegal during the Spring '08 term at Rutgers.

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lecture+7sf - Equilibrium Assume we mix 1.00 mol H2(g) and...

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