Suppose 1.2 moles of hydrazine (N2H4) is added to a 3 L container. At some tempe

Question

Suppose 1.2 moles of hydrazine (N2H4) is added to a 3 L container. At some temperature disproportionation occurs to form nitrogen and hydrogen. At equilibrium, 0.1 moles of hydrogen is present in the container. What are the Kc and Kp values? What is the percent yield? Propose a mechanism for the above reaction. Given the mechanism, does it seem likely that the reaction would proceed quickly or no (use chemical intuition recognizing that each discrete chemical step can be mapped onto a reaction coordinate diagram)?

Explanation / Answer

The dissociation or disproportionation of hydrzine will be

                  N2H4 ---> N2 + 2H2

Initial            1.2          0      0

Change          -x          +x     +2x

Equilibrium    1.2-x       x        2x

Given:

Moles of H2 at equilibrium = 0.1 moles = 2x

[H2] = moles / volume = 0.1/3=0.033 M

Therefore x = 0.05 moles

Moles of N2 = 0.05 moles , [N2] = mole / volume = 0.05 / 3 = 0.0167 M

Moles of N2H4 = 1.2-0.05 = 1.15 moles, [N2H4] = 1.15 / 3 = 0.383 M

Kc = [N2][H2]2 / [N2H4] = 0.0167 X (0.033)2 / 0.383 = 1.017 X 10-4

For Kp we need the pressure or temperature

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