00:02
Given the following equilibrium, we have two no2 gas in equilibrium with 2 no gas and o2 gas.
00:15
And o2 gas for part a, we're asked to write the kc expression, products over reactants, this will be no squared, o2, coefficients become exponents, n02 squared, and there is our kc expression.
00:31
For b, how would the equilibrium be affected by one and increase in temperature? we're told that this is endothermic, so our enthalpy would be on the left -hand side here.
00:49
So if we increase the temperature, then that would result in the heat term increasing, and this would show a shift to the right.
01:01
So an increase in temperature would result in a shift towards the right or towards the products.
01:10
And for change number two, if we increase the total pressure, going back to our equilibrium here, if we increase the pressure, this side here would be our low pressure side as there are two moles of gas.
01:36
This side here would be our high pressure side, as there are three moles of gas.
01:40
Increase the pressure.
01:42
Lachiatide's principle would want to do the opposite.
01:43
It would want to bring the pressure back down to low.
01:46
So this would result in a shift towards the left or shift towards the reactants.
02:01
For part c, given that the nitrogen monoxide was drawn to be 0 .017 molar at equilibrium.
02:13
What was the concentration of nitrogen dioxide in the equilibrium mixture? so we'll set up an ice table here...