MCQ
In a chemical equilibrium $A + B$ $\rightleftharpoons$ $C + D$, when one mole each of the two reactants are mixed, $0.6 $ mole each of the products are formed. The equilibrium constant calculated is
- A$1$
- B$0.36$
- ✓$2.25$
- D$4/9$
Initial $1$ $1$ $0$ $0$
remaining at equilibrium $0.4$ $0.4$ $0.6$ $0.6$
$K = \frac{{[C]\,\,[D]}}{{[A]\,\,[B]}} = \frac{{0.6 \times 0.6}}{{0.4 \times 0.4}} = \frac{{36}}{{16}} = 2.25$.
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${N_2}(g)\, + 3{H_2}(g)\, \rightleftharpoons \,2N{H_3}(g)$
The equilibrium constant of the above reaction is $K_3$. If pure ammonia is left to dissociate, the partial pressure of ammonia at equilibrium is given by (Assume that $P_{NH_3}<\,< P_{total}$ at equilibrium)