MCQ
${P_4} + 3NaOH + 3{H_2}O \to A + 3Na{H_2}P{O_2}$ here, $'A'$ is
- A$N{H_3}$
- ✓$P{H_3}$
- C${H_3}P{O_4}$
- D${H_3}P{O_3}$
${P_4} + 3NaOH + 3{H_2}O \to P{H_3} + 3Na{H_2}P{O_2}$
In this reaction, phosphorus disproportionate into phosphine and sodium hydrogen phosphite.
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Choose the correct answer from the options given below:
$C _{2} H _{6} \rightarrow C _{2} H _{4}+ H _{2}$
the reaction enthalpy $\Delta_{ r } H =...........{ kJ\, mol ^{-1}}$.
(Round off to the Nearest Integer).
[Given : Bond enthalpies in $kJ$ $mol$ $^{-1}:C-C : 347, C = C : 611 ; C - H : 414, H - H : 436]$