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iPad TF3:43 cvg.cengagenow.com ???? Bing Google Yahoo ???? ?+ Write The Cell Notat ????-???? ???????? CengageBrain- My...OWLv2 | Online t.. REVIEW SPRING BREAK Question 1 1 pt For the following reaction at a certain temperature Question 2 H2 (g) + F2 (g)-2HF(g) it is found that equilibrium concentrations in a 8.00-L rigid container are H2] 0.0820 M, F- 0.0350 M, and HF-0.520 M. If0.376 mole of F2 is added to this equilibrium mixture, calculate the concentrations of all gases once equilibrium is reestablished. Question 3 1 pt 1 pt 1 pt 1 pt 1 pt 1 pt 1 pt Question4 Question 5 Concentration of H2 Concentration ofF,- Concentration of HFM Question 6 Question Question 8 Question 9 Submit Answer Try Another Version Question 10 10 item attempts remaining 1 pt 1 pt 1 pt 1 pt 1 pt 1 pt Question 11 Question 12 Question 13 Question 14 Question 15 Question 16 Progress: 0/16 items Due Apr 8 at 11:55 PM Previous Next Finish Assignment Email Instructor Save and Exit Cengage Learning |Cengage Technical SupportExplanation / Answer
Kc = [HF]² / ( [H?]?[F?] )
from the given equilibrium concentrations you can calculate Kc
Kc = (0.52)² / ( 0.082 X 0.035 ) = 94.22
By adding 0.376 mol the fluorine concentrations changes to
[F?]? = 0.035 + 0.376/8 = 0.082 (mol/L)
In order to reestablish equilibrium some fluorine reacts away. Let the decrease of [F?] be x. Due to stoichometry of reaction decrease of
[H?] while [HF] increases by 2x. Hence the equilibrium concentrations are:
[H?] = 0.082 - x
[F?] = 0.082 - x
[HF] = 0.52 + 2?x
Hence:
94.22 = (0.52 + 2?x)² /( (0.082 - x) ? (0.082 - x)
<=>
94.22?(0.082-x)2 = (0.52 + 2?x)²
taking square root of both side and solving it we get,
x = 0.0236
Therefore
[H?] = 0.082 - 0.0236 = 0.0584 M
[F?] = 0.082 - 0.0236 = 0.0584 M
[HF] = 0.52 + 2?0.0236 = 0.5672 M
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