4(c) Consider the reaction 2 NO + O2 <---> 2 NO2 that has a Kc value...

Question

Question

4(c) Consider the reaction 2 NO + O2 <---> 2 NO2 that has a Kc value of 85.5 at temperature Ty. What will happen to the posit

Answer 1

Answer #1

2 NO + O2 = 2 NO2

Qc = (NO2)^2 / ((NO)^2*(O2))

= ( 4.5 )^2 / (( 3.00)^2*(2.50))

= 0.9

Value of Kc = 85.5

Qc value is less than the Kc value so reaction will go forward and equilibrium will shift towards forward reaction

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Answer 2

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