In a reaction mixture containing `H_(2),N_(2)` and `NH_(3)` at partial pressure of 2 atm, 1 atm and 3 atm respectively, the value of `K_(p)` at `725K` is `4.28xx10^(-5)atm^(-2)` . In which direction the net reaction will go ? `N_(2)(g)+3H_(2)(g)hArr 2NH_(3)(g)`
A
Forward
B
Backward
C
No net reaction
D
Direction cannot be predicted
Text Solution
Verified by Experts
The correct Answer is:
2
`Q=(3^(2))/(2^(3)xx1)=(9)/(8)` Since `QgtK_(c)` reaction will proceed in backward direction.
Mass action rato or reaction quotient Q for a reaction can be calculate using the law of masss action A(g)+B(g) hArr C(g)+D(g) Q=([C][D])/([A][B]) The value of Q decides whether the reaction is at equilibrium or not. At equilibrium, Q=K For an equilibrium process, Q ne K when Q gt K , reaction will favour backward direction and when Q lt K , it will favour direction. Answer the following questions: In a reaction mixture containing H_(2), N_(2) and NH_(3) at partial pressure of 2 atm, 1 atm and 3 atm respectively, the value of K_(p) at 700 K is 4.00xx10^(-5) atm^(-2) . In which direction the net reaction will go? N_(2)(g)+3H_(2)(g) hArr 2NH_(3)(g)
A reaction mixture containing H_(2), N_(2) and NH_(3) has partial pressures 2 atm, 1 atm, and 3 at,. Respectively, at 725 K . If the value of K_(p) for the reaction, N_(2)(g)+3H_(2)(g) hArr 2NH_(3)(g) is 4.28xx10^(-5) atm^(-2) at 725 K , in which direction the net reaction will go?
Write the relation between K_(p) " and " K_(c) for the reaction: N_(2)(g) +3H_(2) (g) hArr 2NH_(3)(g)
For the reaction, N_(2)(g)+3H_(2)(g) hArr 2NH_(3)(g) , the units of K_(p) are …………
A reaction mixture containing N_(2) " at " 0*50 " atm" , "at " 0*05 " atm" NH_(3) " and " 3*0 " atm of hydrogen is heated to " 450^(@)C. In which direction the reaction N_(2) (g) + 3 H_(2) (g) hArr 2 NH_(3) (g) " will go if " K_(p)" is" 4*28 xx 10^(-5) ?
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