MCQ
Which of the reaction defines $\Delta H_f^o$ ?
- ✓$\frac{1}{2}{H_2}(g) + \frac{1}{2}{F_2}(g) \to HF(g)$
- B$CO(g) + \frac{1}{2}{O_2}(g) \to C{O_2}(g)$
- C${C_{{\rm{diamond}}}} + {O_2}(g) \to C{O_2}(g)$
- D${N_2}(g) + 3{H_2}(g) \to 2N{H_3}(g)$
Only $(A)$ follows the definition of $\Delta H _{ f }^0$. Therefore, $(A)$ is correct answer.
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Reaction $- (2): CH_3 -CH = CH -CH_3 \xrightarrow{KMn{{O}_{4}}/NaI{{O}_{4}}}(C)$ $2\,mole$
Product $(B)$ and $(C)$ respectively are
(Henry's law constant for $\mathrm{CO}_{2}$ at $298\, \mathrm{~K}$ is $1.67 \times 10^{3}$ $bar$)