The value of Kc for the reaction 3 O21g2 ∆ 2 O31g2 is 1.7 * 10-56...

Question

The value of Kc for the reaction 3 O21g2 ∆ 2 O31g2 is 1.7 * 10-56 at 25°C. Do you expect pure air at 25 °C to contain much O3 (ozone) when O2 and O3 are in equilib- rium? If the equilibrium concentration of O2 in air at 25 °C is 8 * 10-3 M, what is the equilibrium concentration of O3?

Accepted Answer

Hello. Everyone in this video trying to calculate for the equilibrium concentration of N. 02. It's going to be our product in our case. So first things first, we're gonna go ahead and write out what were given to us in this problem where were given to us is going to be the equilibrium constant. That's equal to 0.212 and that's at degrees Celsius. So this equilibrium constant value of 0.212. This value is just around one. So neither of the gasses will be much greater concentration than the other. So basically what I'm trying to say is that at equilibrium the reactant and the products concentrations are almost equal. So that's going to be one of our answers for this problem. The second one is cackling for the concentration of N. 02. Let's first recognize that the equation for this equilibrium constant is equal to well it's a ratio. So for my numerator here it's the concentration of my products, brace the power of its coefficients over the concentration of my reactant, raise your power of its coefficients. So applying it to our chemical equation right over here We have the concentration of N 02 raised to the power of two Over the concentration of N 204. We're trying to isolate this right here because that's what we're solving for. So basically I'm going to do first is multiply both sides by our denominator. So we do So we get that the concentration of N. 02 raised to power of two is equal to our equilibrium constant multiplied by a concentration of our product or our star materials. So that's N. 204. They're finally isolating R. N. 02. Here, we're gonna go and scrub it both sides. So our concentration of N. Two and 02 is equal to, well R. K. C. Multiplied by the concentration of N 204. All raised to power of one over to order square rooted. So I'm gonna go ahead and actually plug in the numerical values for this. So he said that the equilibrium constant value is 0.212. Now for the concentration of N 204 were given this as well, that's 2.0 times 10 to the negative three molar. Again, this is all going to be square rooted. So once I put everything into a calculator, I see that my concentration of N. 02 is equal to 0.206 molar. So this right here is going to be my second and final answer for this problem.

The value of Kc for the reaction 3 O21g2 ∆ 2 O31g2 is 1.7 * 10-56 at 25°C. Do you expect pure air at 25 °C to contain much O3 (ozone) when O2 and O3 are in equilib- rium? If the equilibrium concentration of O2 in air at 25 °C is 8 * 10-3 M, what is the equilibrium concentration of O3?

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