Textbook Problem 5h #3

[805295074]

Use the information in Table 5G.2 to determine the value of K at 300 K for the reaction 2BrCl(g)+H2(g)⇌Br2(g)+2HCl(g).

Can someone help me with this problem?

[Andrew Wang 1C]

Use table 5G.2 to find two or more reactions that add together to result in the desired reaction 2BrCl(g+H2(g)⇌Br2(g)+2HCl(g). Then use the equilibrium constants for the reactions that you found in the table to calculate K for the overall reaction by multiplying them together.

Hope this helps!

[Shrey Pawar 2A]

From the table provided in the book you just have to combine the H2(g)+Cl2(g)⇌2HCl(g) equation with the 2BrCl(g)⇌Br2(g)+Cl2(g) equation. When adding the euqation, the Cl2 with cancel on each side and provide you with the equation you are looking for. In order to get the K you simply multiply the K values together. Hopefully, I explained this well enough!

[Kiran Singh 3A]

First, I would divide up the reaction into two smaller reactions : (1) 2BrCl <--> Br2 + Cl2 and (2) H2 + Cl2 <--> 2 HCl. Using the table, we can determine that the K for (1) is 377 and the K for (2) is 4.0*10^31. To calculate the K for the reaction, we can multiply these two K values together: 377 * 4.0*10^31 = 1.5*10^34. Hope this helps!

[Samantha Pedersen 2K]

If you look at Table 5G.2, you'll see that the reaction 2BrCl(g)+H2(g)⇌Br2(g)+2HCl(g) isn't listed. However, we can get to 2BrCl(g)+H2(g)⇌Br2(g)+2HCl(g) as an overall reaction by "adding" the reactions 2BrCl(g)⇌Br2(g)+Cl2(g) and H2(g)+Cl2(g)⇌2HCl(g) together, and these two reactions are listed in Table 5G.2. Cl2 is present on the right side of the first reaction and the left side of the second reaction, so it cancels out and we are left with the overall reaction 2BrCl(g)+H2(g)⇌Br2(g)+2HCl(g). Whenever we "add" two reactions together to get an overall reaction, we multiply the two values for K together to get the K value for the overall reaction. The K value for 2BrCl(g)⇌Br2(g)+Cl2(g) is 377 and the K value for H2(g)+Cl2(g)⇌2HCl(g) is 4.0 x 10^31. If we multiply 377 x (4.0 x 10^31), we get 1.508 x 10^34 as the value of K for the overall reaction. I hope this helps!

[Hope Fan 2A]

The first step would be finding equations in Table 5.G2 that would result in the equation once added together. A good way to pick out equations is to see whether the equations share reactants or products with the desired equation. The two equations would be: H2(g)+Cl2(g)⇌2HCl(g) and 2BrCl(g)⇌Br2(g)+Cl2(g). You can then find the K values for these equations in Table 5.G2, and then multiply those K values to get the K value of the overall equation.