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Consider the following equilibrium: 2NOCI(g) 2NO(g) + Ch(g) with K = 1.6 × 10 1.00 mole of pure NOC1 and 0.981 mole of pure Cl are placed in a 1.00-L container. Calculate the equilibrium concentration of NO(g). Select one: O a. 9.81 × 10-!M O b. 1.02 M O c.4.04 x 10³M O d. 2.02 x 103 M O e.5.71 x 10 M search

Consider the following equilibrium: 2NOCI(g) 2NO(g) + Ch(g) with K = 1.6 × 10 1.00 mole of pure NOC1 and 0.981 mole of pure Cl are placed in a 1.00-L container. Calculate the equilibrium concentration of NO(g). Select one: O a. 9.81 × 10-!M O b. 1.02 M O c.4.04 x 10³M O d. 2.02 x 103 M O e.5.71 x 10 M search

Chemistry: Principles and Reactions
Chemistry: Principles and Reactions
8th Edition
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:William L. Masterton, Cecile N. Hurley
Chapter12: Gaseous Chemical Equilibrium
Section: Chapter Questions
Problem 13QAP: Consider the following reaction at 250C: A(s)+2B(g)C(s)+2D(g) (a) Write an equilibrium constant...
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Consider the following equilibrium: 2NOCI(g)=2NO(g) + Ch(g) with K = 1.6 × 10°. 1.00 mole of pure NOC1 and 0.981 mole of pure Cl are placed in a 1.00-L container. Calculate the equilibrium concentration of NO(g). tion Select one: O a. 9.81 × 10''M O b. 1.02 M O c. 4.04 × 10-³ M O d. 2.02 × 10³M O e.5.71 x 10 M ere to search F4 F5 F6 EZ F8 F9 F10 F11 F12 Pri %2$ 4. 5 7 E R B N M
Transcribed Image Text:Consider the following equilibrium: 2NOCI(g)=2NO(g) + Ch(g) with K = 1.6 × 10°. 1.00 mole of pure NOC1 and 0.981 mole of pure Cl are placed in a 1.00-L container. Calculate the equilibrium concentration of NO(g). tion Select one: O a. 9.81 × 10''M O b. 1.02 M O c. 4.04 × 10-³ M O d. 2.02 × 10³M O e.5.71 x 10 M ere to search F4 F5 F6 EZ F8 F9 F10 F11 F12 Pri %2$ 4. 5 7 E R B N M
Expert Solution
Step 1

For the equilibrium Reaction -

2NOCl <-----> 2NO + Cl2

The , Equilibrium constant can be written as -

Kc = concentration of Products/concentration of Reactants

Kc = [NO]2[Cl2]/[NOCl]2

Here,

Kc = 1.6×10-5

[NO] =  moles of NO = ?

[Cl2] = moles of Cl2 = 0.981 mole

[NOCl] = moles of NOCl = 1 mole

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