Homework question 5H.1 PART B

Moderators: Chem_Mod, Chem_Admin

Diana Chavez-Carrillo 2L
Posts: 99
Joined: Fri Sep 28, 2018 12:18 am

Homework question 5H.1 PART B

Postby Diana Chavez-Carrillo 2L » Fri Jan 17, 2020 12:21 pm

For the homework problem 5H.1 part B it contains fractions as the stoichiometric coefficient: 1/2 N2 + 3/2 H2 -> NH3. In the solution manual they placed 41 (this number is K=41 which came from part A) to the power of 1/2 and got 6.4 as the value of K for this equation but I am confused about where you get the 1/2 from. Can someone explain the steps to get to (41)^1/2?

MinuChoi
Posts: 80
Joined: Wed Sep 18, 2019 12:15 am

Re: Homework question 5H.1 PART B

Postby MinuChoi » Fri Jan 17, 2020 3:28 pm

The (b) equation is the original equation of the problem scaled by 1/2. If you multiple an entire equation by a scalar (in this case, 1/2), the K constant changes by being raised to that power (in this case, K=41 is raised to the power 1/2 to get K for the part b equation).

805291863
Posts: 80
Joined: Sat Sep 14, 2019 12:16 am

Re: Homework question 5H.1 PART B

Postby 805291863 » Fri Jan 17, 2020 3:31 pm

When multiplying the stoichiometric coefficients of a reaction by x, the K value is raised to the power of x. This is why K=(41)^(1/2)

Also, keep in mind that the K value of the reverse reaction is K^(-1)


Return to “Equilibrium Constants & Calculating Concentrations”

Who is online

Users browsing this forum: No registered users and 2 guests