Consider the reaction. PCl5(g)↽−−⇀PCl3(g)+Cl2(g)Kc=0.0420 PCl 5 ( g ) ↽ − − ⇀ PCl 3 ( g ) + Cl 2 ( g ) K c = 0.0420 The concentrations of the products at equilibrium are [PCl3]=0.270 M [ PCl 3 ] = 0.270 M and [Cl2]=0.180 M [ Cl 2 ] = 0.180 M . What is the concentration of the reactant, PCl5 PCl 5 , at equilibrium? [PCl5]= [ PCl 5 ] =

Generally, for a reaction with low equilibrium constant, the reactant equilibrium will be more favored than those of the products. So in the given problem, we have:

Equilibrium concentration of [tex][PCl_{3}][/tex] = 0.270 M

Equilibrium concentration of [tex][Cl_{2}][/tex] = 0.180 M

The equilibrium constant for the reaction [tex]K_{c}[/tex] = 0.0420

For the given chemical reaction, the balanced equation is:

[tex]PCl_{5(g)}[/tex]⇔[tex]PCl_{3(g)} + Cl_{2(g)}[/tex]

The equation relating the equilibrium constant to the equilibrium concentrations of the reactants and products is:

[tex]K_{c} = \frac{[PCl_{3}][Cl_{2}] }{[PCl_{5}] }[/tex]

Thus:

0.0420 = (0.270*0.180)/[tex][PCl_{5}][/tex]

[tex][PCl_{5}][/tex] = 0.0486/0.042 = 1.157 M

nicomtorres nicomtorres · Answer 2 · 2021-03-19T17:42:42+01:00

nicomtorres nicomtorres

19-03-2021

Answer:

1.16 M

Explanation:

So we know Kc= [PCl3]*[Cl2]/ [PCL5]

Your equation needs to be altered so it gets to the below so we are solving for PCl5 like the problem calls for:

[PCl5]= [PCl3] * [Cl2]/ Kc

Just fill in your given quantities:

[PCl5]= (.180* .270)/.0420

[PCl5]= 1.16 M

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