The equilibrium constant Kc for the gas-phase thermal decomposition of cyclopropane to propene is 1.0 * 105 at 500 K:

(c) Can you alter the ratio of the two concentrations at equilibrium by adding cyclopropane or by decreasing the volume of the container? Explain.

Video transcript

Hello. In this problem we are told the dissociation of iodine to its atoms has an equilibrium constant of 4.6 times 10 -7 at 800 Kelvin. Were asked does the equilibrium ratio of the two concentrations change by adding iodine or reducing the volume of the container. And were asked to justify our answer. So the ratio of the two concentrations that equilibrium is given by our equilibrium crossing expression. So we have the concentration of our products which is atomic iodine squared divided by concentration of our reactant which is molecular iodine. And we're told this is equal to 4.6 times 10 to minus seven At 800 Kelvin. So we are asked to first consider the impact of adding iodine on this ratio. Looking at an equilibrium between molecular iodine and atomic iodine, if we add more iodine then the Shadows principle tells us that the system will respond by moving in the forward direction. And the reason that we'll be moving in the forward direction is because when we add more molecular iodine we are going to increase the rate of the forward reaction relative to the reverse and so it will move in the forward direction. If we look at our ratio of atomic iodine to molecular iodine. What's taking place then, is that the system will take the additional iodine that we've produced and convert it to atomic iodine. Such that this ratio again, once we reach chemical equilibrium is equal to 4.6 times 10 to the -7. So are equal in ratio of the two concentrations remains constant with the addition of iodine. The next thing then we're asked to consider is the impact of volume. So we're told that the volume is reduced according to Boyle's law, pressure and volume are inversely related. So if you reduce the volume the pressure will increase. Principle tells us then that the system will move to side with fewer moles of gas in order to reestablish equilibrium. We look at our equilibrium between molecular iodine and atomic iodine. We have then on the reactive side we have one mole of molecular iodine and we have two moles of atomic iodine. So the system will respond by moving in the reverse direction to reestablish equilibrium. If we make use of the ideal gas equation, we see that the concentration which is moles per unit volume, is equal to the pressure, divided by the gas constant, divided by temperature. So the gas constant temperature are remaining constant. So if the pressure increases then this will correspond to a increase in concentration. And when we decrease the pressure by moving to the side with fewer moles of gas we will decrease the concentration. And so when we look at our equilibrium expression and our ratio of concentration of atomic iodine to like other iodine, the system is again Initially has a higher concentration and it re establishes than our ratio of these two back to 4.6 times 10 to the -7. By moving to the side with fewer moles of gas, this will lower the pressure and then readjust our concentrations so that again, the ratio of our two concentrations remains constant, so the impact on our equilibrium concentration ratio depends on temperature, it doesn't depend on volume. So we saw that when we produce the volume, our equilibrium ratio of the two concentrations remain constant, and when we added iodine, the equilibrium ratio of our two concentrations also remains constant. Thanks for watching. Hope this helps.