So here, this one’s a lot more complicated. What we’re going to say here is we need
to find the KC value for this equation here. EXAMPLE 2: Calculate the rate constant, KC,
for the reaction below: H (g) + Br (g) ---> HBr (g) [KC4]
Use the following information to calculate KC.
H2 (g) ---> 2 H (g) KC1 = 11.8 Br2 (g) ---> 2 Br (g) KC2 =1.15 x 10-5
H2 (g) + Br2 (g) ---> 2 HBr (g) KC3 = 2.78 x 103
So our best strategy for questions like this is since we’re trying to isolate that equation
above, we have to look and see where do we see each one of these elements.
H (g) + Br (g) ---> HBr (g) [KC4] So what we need to do first is find H (g)
and we're going to notice that we see H (g) right here.
H2 (g) ---> 2 H (g) KC1 = 11.8 Now, we have to manipulate this one in red
so it matches up with this one in blue. H (g) + Br (g) ---> HBr (g) [KC4]
H2 (g) ---> 2 H (g) KC1 = 11.8 So what two changes do I need to do to it?
First, the H in red is a product but we need it to be a reactant, so we need to reverse
the reaction. Once we reverse it, we also need to realize that there’s a 2 in front
of this H, but over here in the blue box, there isn’t a 2, so we need to divide by
2 and we just said dividing by 2 is the same thing as multiplying by half. So we have to
reverse it and multiply by a ½. So now our equation becomes:
H (g) ---> ½ H2 (g) When we reverse it, K becomes 1/11.8 and because
we multiplied by ½, half becomes the power and just remember half power really means
square root. H (g) ---> ½ H2 (g) KC1 = (1/11.8)1/2 = √1/11.8
We're going to say that this right here got manipulated so we no longer care about it.
H2 (g) ---> 2 H (g) KC1 = 11.8 Remember, we did all of this so that the red
H (g) that we had matches up with the H (g) in the blue box. We’re basically trying
to manipulate these 3 equations so that they look like the compounds in the blue boxes
above. Next, we’re going to see where do we see
Br (g). H (g) + Br (g) ---> HBr (g) [KC4]
Here in the blue box it’s a reactant but over here it’s a product.
Br2 (g) ---> 2 Br (g) KC2 =1.15 x 10-5 Also here in the blue box, it doesn’t have
a coefficient in front of it, but here it has a 2. So the two changes are the same,
we need to reverse it so it becomes a reactant to and we need to divide by 2 which is the
same thing as multiplying by ½. So this becomes: Br (g) ---> ½ Br2 (g) KC2 =(1/1.15 x 10-5)1/2
= √(1/1.15 x 10-5) Finally HBr, where do we see HBr? Luckily,
we see HBr only one place. We see it right here as a product so they’re both products
so that’s good. So there’s no need to reverse, but there’s a 2 in front of the
HBr, so what do we need to do. We need to divide by 2 or same thing as multiplied by
½ . H (g) + Br (g) ---> HBr (g) [KC4]
H2 (g) + Br2 (g) ---> 2 HBr (g) KC3 = 2.78 x 103
So this is going to become: ½ H2 (g) + ½ Br2 (g) ---> HBr KC3 = (2.78
x 103)1/2 = √2.78 x 103 What we’re going to do next is we’re going
to cross out intermediate, things that look alike but is a product in one equation and
a reactant in another. So we’re going to say here:
H (g) ---> ½ H2 (g) Product
½ H2 (g) + ½ Br2 (g) ---> HBr Reactant
So they’re going to cancel out. Br (g) ---> ½ Br2 (g)
Product ½ H2 (g) + ½ Br2 (g) ---> HBr
Reactant They cancel out. Whatever doesn’t get cancelled
out comes down. So what didn’t get cancelled out?
H (g) + Br (g) ----> HBr (g) The equation we just isolated matches up with
the equation above. Now, all we have to do is we’re going to say KC4 = KC1 x KC2 x
KC3. So basically all we have to do here is we have to multiply each one of these right
here. KC4 = (√1/11.8) x (√1/1.15 x 10-5) x (√2.78
x 103) KC4 = 4526.19
Hope you guys were able to follow along as we did this problem. Just realize, the equation
with the missing KC value is the one we have to look at and we just look compound by compound,
look to see where do we find them in the given equations below. When we’re able to isolate
them, we will just multiply all the K values together and that will give us the overall
K value for this equation we’re looking for in the first place.
So just remember the tips that we did. If you reverse it, K becomes the reciprocal.
If you multiply it by a number, that becomes the power. If you divide by a number, then
it becomes 1 over that number as a power. So if you divide it by 5, it would be multiplying
by 1/5. If you divide it by 4, it would be like multiplying by ¼. Those would become
the powers. Just remember those small, little principles, you’ll be able to get these
questions correct.