Be sure to answer all parts.The equilibrium constant (K) for the formation of nitrosyl chloride, an orange-yellow compound, from nitric oxide and molecular chlorine 2NO(g) + Cl½(g)= 2NOCI(g) is 5 x 10° at a certain temperature. In an experiment, 2.10 × 10 ² mole of NO, 2.10 × 103 mole of Cl,, and 2.80 moles of NOC1 are mixed in a 2.20–L flask. What is Q, for the experiment? x 10 (Enter your answer in scientific notation.) In which direction will the system proceed to reach equilibrium? The reaction will proceed to the left. The reaction will proceed to the right. The reaction is at equilibrium.

Chemistry: The Molecular Science
5th Edition
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:John W. Moore, Conrad L. Stanitski
Chapter12: Chemical Equilibrium
Section: Chapter Questions
Problem 51QRT: At room temperature, the equilibrium constant Kc for the reaction 2 NO(g) ⇌ N2(g) + O2(g) is 1.4 ×...
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Be sure to answer all parts.The equilibrium constant (K) for the formation of nitrosyl chloride, an
orange-yellow compound, from nitric oxide and molecular chlorine
2NO(g) + Cl,(g)= 2NOCI(g)
is 5 x 10° at a certain temperature. In an experiment, 2.10 × 10-2 mole of NO, 2.10 x 10-3 mole of Cl,
and 2.80 moles of NOCI are mixed in a 2.20-L flask.
What is Q, for the experiment?
× 10
(Enter your answer in scientific notation.)
In which direction will the system proceed to reach equilibrium?
The reaction will proceed to the left.
The reaction will proceed to the right.
The reaction is at equilibrium.
Transcribed Image Text:Be sure to answer all parts.The equilibrium constant (K) for the formation of nitrosyl chloride, an orange-yellow compound, from nitric oxide and molecular chlorine 2NO(g) + Cl,(g)= 2NOCI(g) is 5 x 10° at a certain temperature. In an experiment, 2.10 × 10-2 mole of NO, 2.10 x 10-3 mole of Cl, and 2.80 moles of NOCI are mixed in a 2.20-L flask. What is Q, for the experiment? × 10 (Enter your answer in scientific notation.) In which direction will the system proceed to reach equilibrium? The reaction will proceed to the left. The reaction will proceed to the right. The reaction is at equilibrium.
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