Achieve week 1 #5

[305670352]

Can someone please do a step by step for week 1 #5?
I know the rules that follow Hess's law but as there are two products on the right hand side of the given equation I cant figure out how to use it.
If someone is able to help me work through it that would be great, thanks!

[305723807]

saving cuz i need help too

[JacksonWissing2G]

Ok so the first thing we have to do is find the equations that have the products and reactants on the same side as the one we're trying to find. In this case 3*[H2(g)+I2(g)↽−−⇀2HI(g)] gives us the part for HI and 2NH3(g)↽−−⇀N2(g)+3H2(g), which is the inverse of N2(g)+3H2(g)↽−−⇀2NH3(g) gives us the part for N2. Now when finding the adjusted K values of each of these parts we have to cube 160 for H2(g)+I2(g)↽−−⇀2HI(g), as we are using this equation 3 times, and we have to find 1/3.6×10^−2 for N2(g)+3H2(g)↽−−⇀2NH3(g), as we are taking the inverse of this equation. After finding these values we multiply them together and get 1.1*10^8 for the final solution for K.

[Ethan_Choi_2I]

Hi! Jackson did a great job describing this problem. I just want to post a few rules to remember when doing these types of problems. When you invert an equation (like from A+B <-> C+D to C+D <-> A+B) you take the original K value and do 1/K to get the new one. When you want to increase/decrease the coefficients for an equation (lets say A+B <-> C+D to N(A+B <-> C+D) ) then you would take the original K value and do K^N to get the new one. Finally, when adding equations together to find the necessary K value, you would multiply the K values of the constituent equations. Hopefully this makes it easier!

[Mahika Saoji 3K]

Hi! I first started by identifying the reactions with the same products/reactants as the given. These two reactions are: H2+I2 -> 2HI with Ka=160 and N2+3H2->2NH3 with Kc = 3.6 x 10^-2. Once you identify these reactions, we have to make sure that the quantities of the respective reactants/products are equal to the given reaction. In the first equation, we have to multiply the whole thing by 3 to get 6HI. To manipulate the Ka for the first reaction you have to raise it to the power you are multiplying the reaction by (in this case 3), there for the new Ka would be 160^3. In the second reaction, we don't have to change any of the quantities, but we have to manipulate the Kc value because the products and reactants are flipped (in comparison with the given reaction). To do achieve this, we have to do the inverse of Kc. It would be Kc = 1/3.6 x 10^-2. Now that we have these new K values, you multiply them together and get your final answer! That is the key to this problem. I hope that helped!

[Serene Liu 3H]

Hi, so what I did was look at all the equations to see which ones had products that were reactants for another equation that would yield the first given equation that we wanted. So I saw that Equation C in the reverse reaction would use 2NH3 and make N2 and 3H2 (so when using the Kc value I would have to take the reciprocal since I used the inverse of the equation). I also saw that if I multiplied Equation A by 3 I would be get 3H2 + 3I2 yielding 2HI. These two taken together use 2NH3 + 3I2 as reactants and N2 + 6HI as products, which is exactly what we need. So, to get the total K of these two you just multiply 3Ka and 1/Kc. Hope that helps!

[405513470]

why would you multiply the two K values: 1/3.6E-2 and 160^3 instead of adding them?

[sophiavmr]

The rule for finding the equilibrium constant of an overall reaction that is the addition of 2 reactions is that you must multiply the two equilibrium constants of the component reactions (K3 = K1 x K2)

[Saebean Yi 3E]

The first step in this problem is to find which of the given chemical equations allow you to "equal" it to the K equation desired. Remember, you can also reverse or multiply the given equations. If you reverse the equation, you have to take the inverse of the K and if you multiply, you need to take the K to that power (by the number you multiplied by). This might take some time, I usually just write different combinations of equations down and cancel (remember, for example, if there's 2X on the reactant side and 2X on the product side while you calculate the K equation desired, you cancel them out) until I get the answer.

Lastly, all you have to do is multiply, not add, the K's of the equations you used.

[Lexi Tempera 3B]

Hi!

The first and third chemical equations given in the problem are used to calculate the desired equilibrium constant, K.

The first equation is multiplied by 3 and its equilibrium constant, Ka , is cubed. The third equation is reversed and its new equilibrium constant is the inverse of Kc .

3[H2(g)+I2(g)↽−−⇀2HI(g)]
2NH3(g)↽−−⇀N2(g)+3H2(g)
K^3a
1/Kc

[605823283]

this was so helpful! Thank you so much!!!

[305670352]

Thank you so much!