If the value of kc at 25oc is 3.7108, and the equilibrium concentrations for n2 and h2 are 0.000105 m and 0.0000542 m, respectively, use the expression in question 5 to calculate the equilibrium concentration of nh3 .

Equation of decomposition of ammonia:

N2+3H2->2NH3

Euilibrium constant:

Kc=(NH3)^2/((N2)((H2)^3))

As concentration of N2=0.000105, H2=0.0000542

so equation will become:

3.7=(NH3)^2/(0.000105)*(0.0000542)^3

NH3=√(3.7*0.000105*(0.0000542)^3)

NH3=7.8×10⁻⁹

So concentration of ammonia will be 7.8×10⁻⁹.

consider the reaction below.

 no rx

at 500 k, the reaction is at equilibrium with the following concentrations.

[pci5]= 0.0095 m

[pci3] = 0.020

[ci2] = 0.020 m

what is the equilibrium constant for the given reaction?

0.042

0.42

2.4

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