# Equilibrium Constants When Manipulating Equations

Here is some tips and tricks for calculating equilibrium constants, when manipulating chemical equations. So we'll take a look at five cases, and we'll make it as easy as possible for you.

So the first instance is when you reverse an equation. So here we’re given an equation A(gas) yields 2B(gas) and the equivalent constant is given as 5.

So we reverse the equation then we will rewrite it, with products becoming the reactants and the reactants becoming the products. And so what happens is, when you’re reversing an equation, then so you take the inverse of K. And so that means you take the reciprocal just in case.

So in this case, K1' which is the new K would equal 1/5 or 0.2, and so that’s the inverse. So if you reverse the equation, then the products become the reactants and the reactants become the products. So always you flip it. So if you flip the equation, then you take the inverse of K.

K in the second case is if you add equations. So if you add 2 equations and then we have this case that are given. And then if we add them, then we end up with A(g) yields B(g) plus 2c(g), and the reason as to why is because one of those B’s cancels out with one of the B’s from there. You’re left with one B on the product side.

So if you add two equations, then what happens is you multiply the case. So here I have K1 and K2, and so I will get, I will call this K’’. For a lack of a better word. So 5 times 10, the new K will be 50 for this particular reaction, if we add those two equations together.

So for the third one, let’s take a look if you double an equation. So here we have our K1 equation again, and so if I double all the coefficients I get 2A(g) yields 4B(g). So if I double it, then I’ll square K. So K1 is here, so will call this a K1’’’, just to say that all these K’s are different. So I’ll square 5 and I get 25, because 25 is 5 to the second power.

Which brings us to number 4, which is the actually the catcher for number three. So these are the same. So see what happens if I multiply an equation by a number. So I can multiply it by say 3 or ½ or anything like that.

So say if I multiply this 3, then I’ll end up with 3A(g) yields 6B(g). So that means that our K1 will call this K1’’’ equals, or since I multiplied the equation by a number, then you’ll raise K to that number. So in this case it will be five since we multiplied each of the coefficients by 3 will be 5 cubed. And so this K will be equal to 625 then you would have here.

And then the last case is, if you would ever subtract the equations which is very rare, you would end up dividing the K values of each of those equations that you subtracted. So just a recap very easily; reversing an equation, you take the inverse of K. To add two equations together you multiply their K's. You double an equation or multiply an equation by a particular number, then you raise K to that number. If you ever subtract equations you divide the K's. So all this relates because the coefficients become the exponents, and so that’s the reason why that we will take the inverse, or we would take the K to that particular power.

So hopefully this helps break down manipulating chemical equations and figuring out the new equilibrium contents of those other reactions. Have a good one.

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