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Consider the following reaction N2(g) + 3H2(g) 2NH3(g)0.866 atm of N2 and 0.0496 atm of H2 are placed in a flask; when equilibrium is reached, the pressure of NH3 is 0.0310 atm. Calculate Kpfor the reaction. Use the following steps to solvethis problem:

Respuesta :

Answer: [tex]K_p[/tex] is 37928.

Explanation:

Equilibrium constant is the ratio of the concentration of products to the concentration of reactants each term raised to its stochiometric coefficients.

The given balanced equilibrium reaction is,

            [tex]N_2(g)+3H_2(g)\rightarrow 2NH_3(g)[/tex]

at t= 0     0.866 atm     0.0496 atm        0 atm

At eqm. conc.  (0.866-x) atm   (0.0496-3x) atm        2x atm  

The expression for equilibrium constant for this reaction will be,

[tex]K_p=\frac{[p_{NH_3}]^2}{p_{N_2}\times [p_{H_2}]^3}[/tex]

given : 2x = 0.0310

x= 0.0155

Now put all the given values in this expression, we get :   

[tex]K_p=\frac{[2x]^2}{(0.866-x)\times (0.0496-3x)^3}[/tex]

[tex]K_p=37928[/tex]

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