Using Hess's Law To Determine K - Video Tutorials & Practice Problems

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concept

Using Hess's Law To Determine K

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Recall with Hess's law that the enthalpy of reaction or Delta H reaction it changes proportionally to the coefficients of a reaction, meaning that if I have a chemical reaction and I multiply it by 3, I multiply Delta H by 3. If I divide that reaction by 2, I divide Delta H by 2. If I reverse the direction of the reaction I reverse the sign of Delta H. We're going to say however the relationship between your equilibrium constant k and the coefficients of a reaction is exponential. Okay, so when we say exponential we're going to say there were 3 possible rearrangements or changes of a chemical reaction. So here we have our balanced chemical equation, which is 2 moles of sulfur dioxide gas combined with 1 mole of oxygen gas to produce 2 moles of sulfur trioxide gas. Associated with this is our original equilibrium constant value of 71.3. Here we're going to take a look at 3 types of changes that can happen. We can multiply the reaction, we can reverse the reaction, or we could divide the reaction by a value. What effect will this have on my equilibrium constant k? Well, for the first one, multiplication. We're going to say if you multiply the reaction we're going to raise k to the same factor. So for example, I'm going to multiply the reaction by 3. So I'm multiplying this reaction by 3 so all the coefficients get multiplied by 3. I'll have 6SO2 gas plus 3 O2 gas, gives me 6 SO3 gas. When I multiply the reaction by a value, k is raised by that number. So here our new k would be k cubed, so it'd be 71.3 cubed. So that will come out to be pretty large number. It's going to be 362467.097. Here, we don't care about sig figs, per se. We're just seeing how the value is being affected. What happens if I reverse the reaction? Well, if I reverse the reaction then I get the inverse of k. Here this k value is when we have k to the 1, that's it's number. When I reverse it becomes k to the minus one the inverse. So here our new reaction, I'm going to reverse it, it becomes 2 SO 3 gas gives me 2 SO 2 gas plus 1 O 2 gas. I'm just flipping the reaction, my reactants have become products, my product has become a reactant. So now my new case k to the minus 1 which is 71.3 to the minus 1, which is 0.0, 140. That'd be my new k value. Then finally, we could divide the reaction. When you divide the reaction you raise k to a reciprocal of that factor. So here I'm dividing by 2, Dividing by 2 really means I'm raising it to the half power. If I divide it by 3 it'd be k to the 1 third. If I divide it by 4 it become k to the 1 fourth, except. Alright, so here I'm dividing the reaction by 2, so this becomes remember I'm dividing all the coefficients by 2, so this becomes 1 SO2 gas plus half O2 gas gives me 1 SO3 gas. So then it's going to become k to the half power, so that's going to be 71.3 to the half. Remember, half power is the same thing as square root. So I'm taking the square root of 71.3 which gives me an answer of 8.4 4 as my new equilibrium constant. Alright. So remember, the effects that we the changes that we do to our chemical reaction has an exponential change on my equilibrium constant k. It affects the exponent involved with k. So keep this in mind whether you're multiplying, reversing, or dividing your chemical reaction, the changes that it does to your equilibrium constant k.

2

example

Using Hess's Law To Determine K Example

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For the following reaction we're told that the KP value is 7.5 times 10 to the negative 2. Here we need to calculate KP of the reactions below. So basically what's going on here is we have this original equation with its original KP value, and we're changing it either through multiplication, reversal, or division in some way. That will produce a whole new kp value. It's up to us to determine what that new kp value is. If we take a look at the first one in option a, we can see that what is the change that occurred. It looks like we divided this the original equation by 2. Right? So we divided the original equation by 2 so this becomes 1 more. Divide this by 2 this became 1 more. Dividing this by 2 became 2, and divided 1 by 2 gives me a half. Remember, we say that if you're dividing by a number that you're doing the reciprocal of that factor. Basically meaning that I divided by 2 so the new k is going to be k to the half. So here we'd say k p to the half, so that's gonna be 7.5 times 10 to the negative two to the one half power. Remember one half power is the same thing as square root. So this comes out as an answer of point 27 for our new kp value. For the next one, what's the change that we see? Alright, so it looks like our products are now reactants, and it looks like our reactants here are now products. So what did we do? We reversed the reaction. So it's a reversal. When we do a reversal it's going to be the inverse of our original k value, so it's going to be kp to the negative one. So that's 7.5 times 10 to the negative two to the negative one, which equals 13. Here all our answers aren't too sig figs because 7.5 times ten-two has 2 significant figures. And then finally, let's look at the last one. What's the last one? Well, it looks like in the last one, we still reversed it. Our products are now reactants, and our reactants are now products. But what's the other change that occurred? Well, here we had a coefficient of 4, now it's 16. This was a 1, now it's a 4. This these 2 were both twos and now they're eights, so this one will represent a reversal and also we multiply it by 4. Well, when we reverse that's going to give us the reciprocal, but then when you multiply you raise that as a power so it's to the 4. Alright so it's gonna be 7.15 times 10 negative 2 to the negative one, which we know gave us 13, and then we raise that 13 to the 4, so this comes out to 3.2 times 10 to the 4. This would be our final answer for kp. So just remember, we affect these chemical reactions either through multiplication, reversal, or division. Changing the chemical reaction has an exponential change, on the chemical reaction. Okay, so we're going to change the exponent here in order to get our new k value.

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Problem

Problem

K_{c} = 6.5 x 10^{2} at a particular temperature for a reaction: 2 NO(g) + 2 H(g) ⇌ N_{2}(g) + 2 H_{2}O(g). Calculate K_{c} at same temperature for the following reaction: 1/3 N_{2}(g) + 2/3 H_{2}O(g) ⇌ 2/3 NO(g) + 2/3 H(g).

A

8.7

B

0.12

C

0.54

D

3.6×10^{–9}

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