MCQ
de-Broglie equation is
- A$\lambda=\frac{h v}{m}$
- B$\lambda=\frac{m v}{h}$
- C$\lambda= hmv$
- ✓$\lambda=\frac{h}{m v}$
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{C{H_3}{\text{ }} - {\text{ }}CH{\text{ }} = {\text{ }}CH{\text{ }} - {\text{ }}C{H_3}\xrightarrow{{{H_3}{O^ + }}}\,C{H_3} - C{H_2} - CH - C{H_3}}
\end{array}$
$(a)\,\,{N_2}(g) + {O_2}(g) \rightleftharpoons \,\,2NO(g); $ $\Delta {H^o}\, = \,181\,\,kJ$
$(b)\,\,2C{O_2}(g)\,\,\,\, \rightleftharpoons \,2CO(g)\, + \,{O_2}(g);$ $\Delta {H^o}\, = \,566\,\,kJ$
$(c)\,\,{H_2}(g) + {I_2}(g) \rightleftharpoons \,2HI(g) ;$ $\Delta {H^o}\, = \,-9.4\,\,kJ$
$(d)\,\,{H_2}(g) + {F_2}(g) \rightleftharpoons \,2HF(g) ;$ $\Delta {H^o}\, = \,-541\,\,kJ$
${N_2}(g)\, + 3{H_2}(g)\, \rightleftharpoons \,2N{H_3}(g)$
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)