Consider the reaction 3O 2 (g) <----->2O 3 (g) 1.26 moles of O 2 were placed in

Question

Consider the reaction 3O2 (g) <----->2O3 (g) 1.26 moles of O2 were placed in a3.00L flask at 300degrees celcius. There was no O3present intially. At equilibrium, the concentration O2was 0.0824M. Find Kc. "<----->" equals double arrow.
Thanks for your time i will rate!! Consider the reaction 3O2 (g) <----->2O3 (g) 1.26 moles of O2 were placed in a3.00L flask at 300degrees celcius. There was no O3present intially. At equilibrium, the concentration O2was 0.0824M. Find Kc. "<----->" equals double arrow.
Thanks for your time i will rate!!

Explanation / Answer

3O2 2O3 I (initial) 0.42 C (change) -x +x E (equilibrium) 0.0824 x then using algebra you have to solve for x .42-X=0.0824 X=0.3376 Then you use the Kc equation concentration of theproducts at equilibrium divided by the concentration of thereactants at equilibrium but dont forget the stoichiometric part ofthe reaction therefore the concentration of O3 will beto the second power and the concentraton of the O2 willbe to the third power: [0.3376]2 [0.0824]3   = 16.78 3O2 2O3 I (initial) 0.42 C (change) -x +x E (equilibrium) 0.0824 x

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