5) a) Consider the following reaction at equilibrium: respect to partial pressur

Question

5) a) Consider the following reaction at equilibrium: respect to partial pressures of SO, SO, and Oz. (4 points) 1r-49 temperature with Chateli er's principle predicts the effect of an increase in b) Consider the following reaction at equilibrium: (4 points) 2NH3(g) N2(g) + 3H2(g) 1° = +92.4 kJ Le Chatelier's principle predicts that adding N,(g) to the system at eq what changes in the concentration of hydrogen and ammonia equilibrium will result in c) Consider the following reaction at equilibrium: (4 points) C(s) + H20(g)CO(g) + H2(g) What effect will reducing the volume have on the system? (4 points) ) Dinitrogen tetroxide partially decomposes according to the following equilibrium: (5 points) N204(g) 2NO2(g) 1.000-L flask is charged with 8.00x 10-3 mol of N204. At equilibrium, 5.04 x 10-3 mol of 04 remains. Calculate Keq for this reaction.

Explanation / Answer

5.

(a) 2SO2(g)     +     O2(g)     <=====>        2SO3(g)     ; dHo = - 99 kJ

This is an exothermic reaction in forward direction.

Raising the temperature favors the reverse reaction (endothermic)

Lowering the temperature favors the forward reaction (exothermic)

So, the partial pressure of SO3(g) will decrease and that of SO2(g) and O2(g) will increase.

(b)

2NH3(g)     <=====>        N2(g)     +      3H2(g)        ; dHo = + 92.4 kJ

According to Le Chatelier, the position of equilibrium will move in such a way as to counteract the change. That means that the position of equilibrium will move so that the concentration of N2(g) decreases again - by reacting it with H2(g) and turning it into NH3. The position of equilibrium moves to the left.

So, the concentration of H2(g) will decrease and that of NH3(g) will increase.

(c)

C(s)     +     H2O(g)     <=====>        CO(g)     +      H2(g)       

When there is a decrease in volume, the equilibrium will shift to favor the direction that produces fewer moles of gas.

When there is an increase in volume, the equilibrium will shift to favor the direction that produces more moles of gas.

So, the concentration of H2O(g) will increase.

(d)

                      N2O4(g)     <=====>        2NO2(g)
IC:                  0.008                                      0
C:                     - x                                     + 2x
EC:              0.00504                                    2x

So, x= 0.00504

Keq = [NO2]2 / [N2O4]
       = (2x)2 / x
       = 4x2 / x
       = 4x
       = 4 (0.00504)
       = 0.02016

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