Respuesta :
Answer: The equilibrium constant for the total reaction is [tex]4.09\times 10^{-6}[/tex]
Explanation:
We are given:
[tex]K_{c_1}=0.282\\\\K_{c_2}=41[/tex]
We are given two intermediate equations:
Equation 1: [tex]N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g);K_{c_1}=0.282[/tex]
The expression of [tex]K_{c_1}[/tex] for the above equation is:
[tex]K_{c_1}=\frac{[NH_3]^2}{[N_2][H_2]^3}[/tex]
[tex]0.282=\frac{[NH_3]^2}{[N_2][H_2]^3}[/tex] .......(1)
Equation 2: [tex]H_2(g)+I_2(g)\rightleftharpoons 2HI(g);K_{c_2}=41[/tex]
The expression of [tex]K_{c_2}[/tex] for the above equation is:
[tex]K_{c_2}=\frac{[HI]^2}{[H_2][I_2]}[/tex]
[tex]41=\frac{[HI]^2}{[H_2][I_2]}[/tex] ......(2)
Cubing both the sides of equation 2, because we need 3 moles of HI in the main expression if equilibrium constant.
[tex](41)^3=\frac{[HI]^6}{[H_2]^3[I_2]^3}[/tex]
Now, dividing expression 1 by expression 2, we get:
[tex]\frac{K_{c_1}}{K_{c_2}}=\left(\frac{\frac{[NH_3]^2}{[N_2][H_2]^3}}{\frac{[HI]^6}{[H_2]^3[l_2]^3}}\right)\\\\\\\frac{0.282}{68921}=\frac{[NH_3]^2[I_2]^3}{[N_2][HI]^6}[/tex]
[tex]\frac{[NH_3]^2[I_2]^3}{[N_2][HI]^6}=4.09\times 10^{-6}[/tex]
The above expression is the expression for equilibrium constant of the total equation, which is:
[tex]2NH_3(g)+3I_2(g)\rightleftharpoons 6HI(g)+N_2;K_c[/tex]
Hence, the equilibrium constant for the total reaction is [tex]4.09\times 10^{-6}[/tex]
