How much faster will a reaction proceed if the temperature is increased by 55.2
ID: 1021534 • Letter: H
Question
How much faster will a reaction proceed if the temperature is increased by 55.2 oC. Given: A + B doubleheadarrow C + 2 D 1 mole of A and 1 mole of B are placed in a 5.0 L container. After equilibrium has been reached, .25 mole of C is present in the container. Calculate the equilibrium constant for the reaction. Consider the reaction at equilibrium N 2 + O 2 + heat doubleheadarrow 2 NO What will happen to the concentration of NO at equilibrium if A catalysts is added more O 2 is added For the reaction A + B doubleheadarrow C + D The equilibrium constant is 100 at a certain temperature If 0.400 mole each of A and B are placed in a 2.0 L container at the temperature, what is the concentration of B at equilibrium?Explanation / Answer
7. Every 10C0 rise rate reaction will be double. 25 = 32 times faster
8. A + B -------- >C + 2D
I 1 1 0 0
C -0.25 -0.25 0.25 2*0.25
E 1-0.25 1-0.25 0.25 0.5
0.75 0.75 0.25 0.5
[A] = 0.75/5 = 0.15M
[B] = 0.75/5 = 0.15M
[C] = 0.25/5 = 0.05M
[D] = 0.5/5 = 0.1M
Kc = [C][D]2/[A][B]
= 0.05*(0.1)2/(0.15)2
= 0.022
9. add catalyst conc of NO is no change because of catalyst increase both farward and back ward reaction.
b. added O2 increases the conc of NO
conc of reactants increases equilibrium favor farward direction.
10. A + B --------> C + D
I 0.4 0.4 0 0
C -x -x +x +x
E 0.4-x 0.4-x +x +x
[A] = 0.4-x/2
[B] = 0.4-x/2
[C] = x/2
[D] = x/2
Kc = [C]{D]/[A][B]
100 = x*x/0.4-x*0.4-x
10 = x/0.4-x
x = 10*(0.4-x)
x = 0.36
[B] = 0.4-x/2 = 0.4-0.36/2 = 0.02M
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