A mixture of 1.57 mol of N_2 1.92 mol of H_2 and 8.13 mol of NH_3 is introduced into a 20 L reaction vessel at 500 K. At this temperature, the equilibrium constant, K_c for the reaction N_(2(g)) + 3H_(2(g)) hArr 2NH_(3(g)) is 1.7 xx 10^2. Is the reaction mixture at equilibrium? If not, what is the direction of the net reaction ?

A mixture of 1.57 mol of N_2 1.92 mol of H_2 and 8.13 mol of NH_3 is i

1.	A mixture of 1.57 mol of N_2 1.92 mol of H_2 and 8.13 mol of NH_3 is introduced into a 20 L reaction vessel at 500 K. At this temperature, the equilibrium constant, K_c for the reaction N_(2(g)) + 3H_(2(g)) hArr 2NH_(3(g)) is 1.7 xx 10^2. Is the reaction mixture at equilibrium? If not, what is the direction of the net reaction ?
Answer» <html><body><p></p>Solution :`{:("<a href="https://interviewquestions.tuteehub.com/tag/equilibrium-974342" style="font-weight:bold;" target="_blank" title="Click to know more about EQUILIBRIUM">EQUILIBRIUM</a> reaction :",N_(2(g)) +, 3H_(2(g)) <a href="https://interviewquestions.tuteehub.com/tag/harr-2692945" style="font-weight:bold;" target="_blank" title="Click to know more about HARR">HARR</a> , 2NH_(3(g))),("Mol of mixture :",1.57,1.92,8.13),("mol" L^(-1) "of equili. mixture :", 1.57/20,1.92/20,8.13/20),(,0.0785,0.096,0.4065):}` <br/> The equilibrium reaction mixture quotient, `Q_c` <br/> `therefore Q_c=[NH_3]^2/([N_2][H_2]^3)=(0.4065)^2/((0.0785)(0.096)^3)` <br/> Here, `(Q_c=2.3792) lt (K_c=1.7xx10^2)` . The <a href="https://interviewquestions.tuteehub.com/tag/value-1442530" style="font-weight:bold;" target="_blank" title="Click to know more about VALUE">VALUE</a> of `Q_c` is less . So, the reaction will proceed in the direction of the products (<a href="https://interviewquestions.tuteehub.com/tag/forward-464460" style="font-weight:bold;" target="_blank" title="Click to know more about FORWARD">FORWARD</a> reaction).</body></html>