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Quotient Rule of Differentiation

For BrO3-⟶ ...

Question

For $B r O_{3}^{-} ⟶ B r^{-}$ , the relation between normality and molarity is given as $N = M \times x$ , value of $x$ is:

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Solution

$B r O_{3}^{-} ⟶ B r^{-}$
+5 -1
As we know,
Normality $=$ Molarity $\times n -$ factor $= M \times 6$ .
So, value of $x$ is $6$ .

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A s O_{2}^{2 -} + B r O_{3}^{-} + H^{+} ⟶ B r^{-} + A s O_{4}^{-} + H_{2} O

Equivalent weight of

B r O_{3}^{-} = \frac{M}{x}

x

Q. In acidic medium, the rate of reaction between

B r O_{3}^{-}

B r^{-}

is given by the expression

- \frac{d [B r O_{3}^{-}]}{d t} = k [B r O_{3}^{-}] [B r^{-}] [H^{+}]^{2}

Q. In acidic medium, the rate of reaction between

B r O_{3}^{⊖}

B r^{⊖}

is given by the expression

\frac{d [B r O_{3}^{⊖}]}{d t} = k [B r O_{3}^{⊖}] [B r^{⊖}] [H^{\oplus}]^{2}

Select correct statements.

Q. For the reaction,

B r O_{3}^{-} (a q) + 5 B r^{-} (a q) + 6 H^{+} \to 3 B r_{2} (l) + 3 H_{2} O (l)

, the rate equation can be expressed in two ways

\frac{d [B r_{2}]}{d t} = k [B r^{-}] [B r O_{3}] [H^{+}]^{2}

- \frac{d [B r^{-}]}{d t} = k^{'} [B r^{-}] [B r O_{3}] [H^{+}]^{2}

Here, k and k' are related as

Q. In acidic medium the rate of reaction between

(B r O_{3})^{-}

B r^{-}

ion is given by the expression.

- [d (B r O_{3}^{-}) / d t] = K [B r O_{3}^{-}] [B r^{-}] [H^{+}]^{2}

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