If 0.40 moles of PCl5 is heated in a 10.0 L container,...

If 0.40 moles of PCl5 is heated in a 10.0 L container, equilibrium is established in which 0.25 moles of Cl2 are present. The reaction is: PCl5(g) ? PCl3(g) + Cl2(g) What is the value of the equilibrium constant?

Transcript

00:01 Hi, we are to solve for the equilibrium constant of the given reaction here.

00:06 So let me just rewrite it.

00:08 We have pcl5, this is our reactant.

00:12 And then on the product side, we have pcl3 and chlorine gas.

00:20 Now first, we need to write our i's table, initial concentration, change in concentration, and equilibrium concentration.

00:28 Concentration we have here the initial amount of the reactant pcl5 0 .40 moles and this is concentration so that means we have to divide the number of moles by the volume which is 10 liters and the products are zero initially so the change this will lose some amounts since this is the reactant and this will gain plus x plus x at the equilibrium we'll have initial minus change so 0 .4 d minus 10 oh sorry divided by 10 rather 0 .4 divided by 10 is 0 .04 and then minus x this is a change this is equivalent to x of equilibrium products it's also x and it is stated here that um at the equilibrium we have 0 .25 moles of chlorine gas present so that means x is equivalent to 0 .25 over the volume, which is 10 liters.

01:29 0 .25 divided by 10 is 0 .025 molar or moles per liter.

01:38 So this is the value of x.

01:40 So this x is also 0 .025.

01:44 And 0 .04 minus the value of x, 0 .025.

01:49 This will give us 0 .015.

01:53 So now that we know the concentration at equilibrium of the involved compounds, you can now solve for the equilibrium constant kc.

02:00 This is equivalent to the concentration of the products at equilibrium...

Question

If 0.40 moles of PCl5 is heated in a 10.0 L container, equilibrium is established in which 0.25 moles of Cl2 are present. The reaction is: PCl5(g) ? PCl3(g) + Cl2(g) What is the value of the equilibrium constant?

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