MCQ
What is the de-Broglie wavelength associated with the hydrogen electron in its third orbit
- A$9.96 \times {10^{ - 10}}\,cm$
- ✓$9.96 \times {10^{ - 8}}\,cm$
- C$9.96 \times {10^4}\,cm$
- D$9.96 \times {10^8}\,cm$
Since,
$\frac{ nh }{2 \pi}= mvr$
$\frac{ h }{ mv }=\frac{2 \pi r }{ n }$
$\lambda=\frac{ h }{ mv }=\frac{2 \pi \times 9 R }{3}$
$\lambda=6 \pi R$
We know that
$R =0.529\, \mathring A$
Therefore,
$\lambda=6 \pi \times 0.529$
$\lambda=9.96\, \mathring A$
$\lambda=9.96 \times 10^{-8} \,cm$
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$(i)\,CO(g)+ H_2O(g) \rightleftharpoons CO_2(g)+H_2(g)\,;\,K_1$
$(ii)\,CH_4(g)+H_2O(g) \rightleftharpoons CO(g)+3H_2(g)\,;\,K_2$
$(iii)\,CH_4(g) + 2H_2O(g) \rightleftharpoons CO_2(g)+ 4H_2(g)\,;\,K_3$
Propene $\xrightarrow{{C{l_2}/500\,{}^oC}}A\xrightarrow[{{\text{dryether}}}]{{Na}}B$
$B$ is
$\left( i \right)\,2F{e_2}{O_3}\left( s \right) \to 4Fe\left( s \right) + 3{O_2}\left( g \right)$
${\Delta _r}{G^o} = + 1487.0\,kJ\,mo{l^{ - 1}}$
$\left( {ii} \right)\,2CO\left( g \right) + {O_2}(g) \to 2C{O_2}\left( g \right)$
${\Delta _r}{G^o} = - 514.4\,kJ\,mo{l^{ - 1}}$
Free energy change, $\Delta_rG^o$ for the reaction
$\,2F{e_2}{O_3}\left( s \right) + 6CO\left( g \right) \to 4Fe\left( s \right) + 6C{O_2}\left( g \right)$ will be .....$kJ\, mol^{-1}$