# Question 1a1ea

Nov 20, 2016

The equilibrium constant, ${K}_{\text{c}}$, for any reversible reaction, $\text{aA"+"bB"rightleftharpoons "cC"+"dD}$, is $\left(\left[\text{C"]^("c")["D"]^("d"))/(["A"]^("a")["B"]^("b}\right)\right)$

#### Explanation:

That is, the equilibrium constant equals the concentration of the products raised to the power of their moles divided by the concentration of the reagents raised to the power of their moles.

In this case, ${K}_{\text{c"=((0.430)^3(0.800)^1)/0.250^2=1.102 " mol"^2 " dm}}^{-} 6$

As for working out the units for ${K}_{\text{c}}$ we simply replace the actual concentrations with the units for concentration:

K_"c"==((cancel("mol " "dm"^-3))(cancel("mol " "dm"^-3))("mol " "dm"^-3)("mol " "dm"^-3))/((cancel("mol " "dm"^-3))(cancel("mol " "dm"^-3))1#

${K}_{\text{c"=("mol " "dm"^-3)("mol " "dm"^-3)="mol"^2 " dm}}^{-} 6$