At a given temperature, the equilibrium constant for reaction PC l_5 _ (g) PC l_3 _ (g) + C l_2 _ (g) is 2.4 10^ - 3 . At the same temperature, the equilibrium constant for reaction PC l_3 _ (g) + C l_2 _ (g) PC l_5 _ (g) is

At a given temperature, the equilibrium constant for reaction PC …

MCQ

At a given temperature, the equilibrium constant for reaction $PC{l_5}_{(g)}$ $\rightleftharpoons$ $PC{l_3}_{(g)} + C{l_2}_{(g)}$ is $2.4 \times {10^{ - 3}}$. At the same temperature, the equilibrium constant for reaction $PC{l_3}_{(g)} + C{l_2}_{(g)}$ $\rightleftharpoons$ $PC{l_5}_{(g)}$ is

A
$2.4 \times {10^{ - 3}}$
B
$ - 2.4 \times {10^{ - 3}}$
✓
$4.2 \times {10^2}$
D
$4.8 \times {10^{ - 2}}$

✓

Answer

Correct option: C.

$4.2 \times {10^2}$

$K = \frac{1}{{(2.4 \times {{10}^{ - 3}})}} = 4.2 \times {10^2}$

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$\mathrm{RCONH}_2$ is converted into $\mathrm{RNH}_2$ by means of Hofmann bromamide degradation.

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$1.$ How can the conversion of $(i)$ to $(ii)$ be brought about?

$(A)$ $\mathrm{KBr}$ $(B)$ $\mathrm{KBr}+\mathrm{CH}_3 \mathrm{ONa}$ $(C)$ $\mathrm{KBr}+\mathrm{KOH}$ $(D)$ $\mathrm{Br}_2+\mathrm{KOH}$

$2.$ Which is the rate determining step in Hofmann bromamide degradation?

$(A)$ Formation of $(i)$ $(B)$ Formation of $(ii)$ $(C)$ Formation of $(iii)$ $(D)$ Formation of $(iv)$

$3.$ What are the constituent amines formed when the mixture of $(i)$ and $(ii)$ undergoes Hofmann bromamide degradation?

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give the answer question $1$, $2$, and $3.$