## Homework problem 5I.11

LeAirraBullingor2k
Posts: 77
Joined: Wed Mar 13, 2019 12:15 am

### Homework problem 5I.11

Can anyone help me to understand how they got 6.9 as the answer to the first part of the question?

Pegah Nasseri 1K
Posts: 100
Joined: Wed Feb 27, 2019 12:15 am

### Re: Homework problem 5I.11

For the first part of this question, you are calculating Q which is the concentration of the products over the concentration of the reactants (not at equilibrium). Use the moles of the reactants and products given to you in the question and divide each one by the volume (0.500 L) to obtain a concentration for each reactant and product. Make sure the equation is balanced: 2SO2 (g) + O2 (g) -> 2SO3 (g). The Q is calculated as follows: Q = [SO3]^2 / ([O2][SO2]^2).

Miriam Villarreal 1J
Posts: 105
Joined: Sat Aug 17, 2019 12:16 am

### Re: Homework problem 5I.11

1.20x(0.001) or 10^-3/.500L= SO2
0.50x(0.001) or 10^-3/.500L=O2
0.10x(0.001) or 10^-3/.500=SO3

set up the reaction quotient to calculate Qc (product/reactant)
(0.0002)^2
_________________ = 6.94
(.001) x (.0024)^2

since this value is more than the provided Kc, more product (SO3)will form shifting to the right

Jamie Lee 1F
Posts: 106
Joined: Fri Aug 09, 2019 12:16 am

### Re: Homework problem 5I.11

In order to get Qc, you have to us [products]/[reactants. However, in this problem, you have to change mmol to mol, and put that value of moles over 0.5 L to find M, or Molarity. Once you use these conversions, [SO3]^2/[SO2]^2[O2] should give you Qc=6.9.