Given : MnO 2 + 4 HCl → MnCl 2 + 2 H 2 O + Cl 2 The rate of disappearance of HCl is 8 × 1 0 − 4 m o l l i t r e − 1 mi n − 1 . The rate of appearance of Cl 2 is

4 ( RO D ) H Cl = ( RO A ) C l 2 ( RO A ) C l 2 = 4 8 × 1 0 − 4 = 2 × 1 0 − 4 M mi n − 1 .

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Given : $MnO_{2} + 4 HCl \to MnCl_{2} + 2 H_{2} O + Cl_{2}$

The rate of disappearance of HCl is $8 \times 1 0^{- 4} m o l l i t r e^{- 1} mi n^{- 1}$ . The rate of appearance of $Cl_{2}$ is

Answer:

Solution:

$\frac{( RO D ) _{H Cl}}{4} = (RO A)_{C l_{2}}$

$(RO A)_{C l_{2}} = \frac{8 \times 1 0 ^{- 4}}{4} = 2 \times 1 0^{- 4} M mi n^{- 1} .$

Stream:JEESubject:ChemistryTopic:Chemical KineticsSubtopic:Rate of Reactions/Rate Law

⏱ 2mℹ️ Source: QB

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