Consider the following equilibrium: 2NOCI(g) 2NO(g) + Cl 2(g) with K = 1.6 \' 10

Question

Consider the following equilibrium: 2NOCI(g) 2NO(g) + Cl 2(g) with K = 1.6 ' 10"5. 1.00 mole of pure NOCI and 0.958 mole of pure Cl2 are placed in a 1.00-L container. Calculate the equilibrium concentration of NO(g). 2.04 '10-3M 5.78 '10-3M 1.04 M 4.09 ' 10-3/W 9.58 ' 10-1 M Points Earned: 1.0/1.0 The following questions refer to the equilibrium shown here: 4NH3(g) + s02(g) 4NO(g) + 6H20(g) What would happen to the system if oxygen were added? More oxygen would be produced. Nothing would happen. More monia would be produced. The equilibrium would shift to the right. The equilibrium would shift to the left.

Explanation / Answer

1) Write equilibrium constant equation & substitute the values for concentrations of reactant and products at equilibrium , assuming that x moles of NOCl are used up to form 2x moles of NO and x moles of Cl2 as per stoichiometry of the reaction-

2 NOCl(g)     < --------------------> 2NO(g) + Cl2 (g)     

given initial concentrations 0.1 mole 0.958 moles   

Concentration at equilibrium (1.00 - x ) 2x ( x + 0.958 )

Now , since the volume of container is 1.0 litre-

At equilibrium - [ NOCl(g) ] = ( 1.00 - x )M

[ NO(g)] = 2x M

[ Cl2(g) ] = ( 0.958 + x ) M {Because 0.958 moles of Cl2  is already present in the reaction container .}

Now, K = ( [ NO(9) ]2 [ Cl2(g)    ] ) / [ NOCl (g) ]2        

1.6 X 10-5 = ( 2x )2 ( 0.95 8 + x ) / ( 1.00 - x )2

Solve the equation to get the value of x , 2x = 4.09 X 10-3 M as correct answer , representing the concentration of NO(g )   at this equilibria.

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According to Le - Chatelier's principle addition of any one of the reactants ( in this case oxygen ) would shift the equilibria to a direction where it gets utilized . So equilibrium would shift to right yielding more ammonia.

Answer D ) The equilibrium would shift to right.

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