C www-awu.aleks.com/alekscgi/x/lsl.exe/10_u-IgNslkr7j8P3jH-IvTqeviKFP6W0cqJcWJdIACROQwyw24GWHInrCnyqVCaf_xSaZNy Solubility and... 18.3 Gibbs Free E... 5.3 Enthalpies of... 18.5 Gibbs Free E... Reading Schedule 19.6 Reduction Po... Y SOLUTI E O KINETICS AND EQUILIBRIUM Calculating equilibrium composition from an equilibrium constant Suppose a 500. mL flask is filled with 1.4 mol of NO3 and 1.5 mol of NO₂. The following reaction becomes possible: NO₂(g) + NO(g) 2NO₂(g) The equilibrium constant K for this reaction is 3.89 at the temperature of the flask. Calculate the equilibrium molarity of NO₂. Round your answer to two decimal places. M 1 X S

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Solubility and...
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OKINETICS AND EQUILIBRIUM
Calculating equilibrium composition from an equilibrium constant
M
Explanation
Suppose a 500. mL flask is filled with 1.4 mol of NO3 and 1.5 mol of NO₂. The following reaction becomes possible:
NO₂(g) + NO(g) 2NO₂(g)
The equilibrium constant K for this reaction is 3.89 at the temperature of the flask.
Calculate the equilibrium molarity of NO₂. Round your answer to two decimal places.
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Transcribed Image Text:C www-awu.aleks.com/alekscgi/x/lsl.exe/10_u-IgNslkr7j8P3jH-IvTqeviKFP6W0cqJcWJdIACROQwyw24GWHInrCnyqVCaf_xSaZNy Solubility and... 18.5 Gibbs Free E... esc ||| 18.3 Gibbs Free E... OKINETICS AND EQUILIBRIUM Calculating equilibrium composition from an equilibrium constant M Explanation Suppose a 500. mL flask is filled with 1.4 mol of NO3 and 1.5 mol of NO₂. The following reaction becomes possible: NO₂(g) + NO(g) 2NO₂(g) The equilibrium constant K for this reaction is 3.89 at the temperature of the flask. Calculate the equilibrium molarity of NO₂. Round your answer to two decimal places. @ 2 5.3 Enthalpies of... Check W # 3 X E 10 54 $ S > R C % 5 e T Reading Schedule tv MacBook Pro ^ 6 & 19.6 Reduction Po... SOLUTI 7 Ⓒ2022 McGraw Hill LLC. All Rights Reser U * 00 8
Expert Solution
Step 1

Given -

NO3(g) +NO(g) <-> 2NO2(g)

Volume = 500 mL = 0.500 L (1 L = 1000ml) 

Mole of NO3= 1.4 mol

Mole of NO2= 1.5 mol 

K=3.89

 

 

 

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