Equilibrium Constant Value
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GIVEN THE EQUILIBRIUM CONSTANT VALUE OF Kc FOR THE FOLLOWING RECTION AT 300 CELSIUS
N2(g) + 1/2 O2(g) N2O(g) Kc=3.4 * 10 XY -18(SQUARE ROOT) AT 300 CELSIUS
N2O4(g) 2NO2(g) kc=4.6*10(SQUARE RROT OF -3 AT 300 CELSIUS
1/2N2(g) + O2(g) NO2(g) kc=4.1*10xy -9 at 300 CELSIUS
1- WHAT IS THE VALUE OF Kc FOR THE FOLLOWING REACTION AT 300 CELSIUS.
2N2O(g) + 3O2(g) 2N2O4(g)
2- WHAT IS THE VALUE OF Kp FOR THE REACTION IN EXERCISE 1?
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First, let's write each equilibrium down.
N2(g) + 1/2 O2(g) <----------> N2O(g)
N2O4(g) <-----------> 2NO2(g)
1/2 N2(g) + O2(g) <-------> NO2(g)
Next, in order to arrive at the unknown equilibrium, 2 N2O(g) + 3 O2(g) <--------> 2 N2O4(g), we must multiply the first reaction by two and reverse it, multiply the second reaction by 2 and reverse it, and multiply the third reaction by 4. Like so,
2 N2O(g) <----------> 2 N2(g) + O2(g)
4 NO2(g) <-----------> 2 N2O4(g)
2 N2(g) + 4 O2(g) <-------> 4 NO2(g)
Now, we can add them up to arrive at the reaction of interest:
2N2O(g) + 3O2(g) <--------> 2 N2O4(g)
Therefore, for each reaction, we must change Kc accordingly. Since we multiplied the equation for the reaction by a number, n, we must raise the original value of Kc to the nth power. ...
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