Homework problem 5I.11

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Homework problem 5I.11

Postby LeAirraBullingor2k » Thu Jan 23, 2020 5:59 pm

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

Pegah Nasseri 1K
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Re: Homework problem 5I.11

Postby Pegah Nasseri 1K » Thu Jan 23, 2020 6:23 pm

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
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Re: Homework problem 5I.11

Postby Miriam Villarreal 1J » Fri Jan 24, 2020 2:32 pm

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)
_________________ = 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
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Re: Homework problem 5I.11

Postby Jamie Lee 1F » Sat Jan 25, 2020 5:13 pm

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.

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