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    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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    https://brainmass.com/chemistry/chemical-equilibrium/equilibrium-constant-value-40081

    Solution Preview

    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. ...

    Solution Summary

    The question is solved step-by-step with each step explained in detail.

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