A mixture of1*57 " mol of " N_(2) , 1*92 " mol of " H_(2) and8*13 " mol of " NH_(3) " is introduced into a20 L reaction vessel at 500 K. At this temperature , the equilibrium constant ," K_(c) for the reaction , N_(2) (g) + 3 H_(2) (g) hArr 2 NH_(3) " is " 1*7 xx 10^(2) Is the reaction mixture at equilibrium ? If not, what is the direction of the net reaction ?

A mixture of1*57 " mol of " N_(2) , 1*92 " mol of " H_(2) and8*13 " mo

1.	A mixture of157 " mol of " N_(2) , 192 " mol of " H_(2) and813 " mol of " NH_(3) " is introduced into a20 L reaction vessel at 500 K. At this temperature , the equilibrium constant ," K_(c) for the reaction , N_(2) (g) + 3 H_(2) (g) hArr 2 NH_(3) " is " 17 xx 10^(2) Is the reaction mixture at equilibrium ? If not, what is the direction of the net reaction ?
Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :The reaction is : ` N_(2) (g) + 3 H_(2) (g) hArr 2 NH_(3) (g) ` <br/> ` Q_(c) = ([NH_(3)]^(2))/([N_(2) ][H_(2)]^(3)) = ((8<a href="https://interviewquestions.tuteehub.com/tag/13-271882" style="font-weight:bold;" target="_blank" title="Click to know more about 13">13</a>)/<a href="https://interviewquestions.tuteehub.com/tag/20-287209" style="font-weight:bold;" target="_blank" title="Click to know more about 20">20</a>"mol"L^(-1))^(2)/(((157)/2"mol" L^(-1))((192 )/20 "mol" L^(-1))^(3))=2 <a href="https://interviewquestions.tuteehub.com/tag/38-310079" style="font-weight:bold;" target="_blank" title="Click to know more about 38">38</a> xx 10^(3)` <br/>As ` Q_(c) != K_(c),` the reaction mixture is not in equilibrium .<br/> As `Q_(c) gtK_(c)`, the et reaction will be in the backward direction .</body></html>