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Standard XII

Chemistry

Types of Elementary Reactions

For a consecu...

Question

For a consecutive reaction $R_{1} k_{1} - \to R_{2} R_{1} k_{2} - \to R_{3}$ If initial concentration of $R_{1}$ is 100 M and $k_{1} : k_{2} = 1 : 0.15$ the value of $[R_{2}]_{m a x}$ is $[G i v e n : k_{1} = 4.0 \times 10^{- 2} m i n^{- 1}]$

71.55 M

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55.71 M

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7.155 M

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5.571 M

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Solution

The correct option is A
71.55 M

The values of $[R_{2}]_{m a x}$ is given by the relation
$[R_{2}]_{m a x} = [R_{1}]_{0} [\frac{k_{2}}{k_{1}}]^{\frac{k_{2}}{(k_{2} - k_{1})}}$
$∴ [R_{2}]_{m a x} = 100 [\frac{0.15 \times 10^{- 2}}{1.0 \times 10^{- 2}}]^{\frac{0.015 \times 10^{- 2}}{(1 - 0.15) \times 10^{- 2}}} = 100 [\frac{0.15}{1}]^{\frac{0.15}{0.85}} = 71.55 M .$

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For a consecutive reaction $R_{1} k_{1} - \to R_{2} R_{1} k_{2} - \to R_{3}$ If initial concentration of $R_{1}$ is 100 M and $k_{1} : k_{2} = 1 : 0.15$ the value of $[R_{2}]_{m a x}$ is $[G i v e n : k_{1} = 4.0 \times 10^{- 2} m i n^{- 1}]$

Q. Two reactions

R_{1}

and

R_{2}

have identical pre-exponential factors. Activation energy of

R_{1}

exceeds that of

R_{2}

10 K j m o l^{- 1}

K_{1}

and

K_{2}

are rate constants for reactions

R_{1}

and

R_{2}

respectively at

300 K

, then in

(K_{2} / K_{1})

is equal to:

(R = 8.314 j m o l^{- 1} K^{- 1})

Q. Two reactions

R_{1}

and

R_{2}

have identical pre-exponential factors. Actiation energy of

R_{1}

exceeds that of

R_{2}

10 k J^{- 1}

. If

K_{1}

and

K_{2}

are rate constants for reactions

R_{1}

and

R_{1}

respectively at 300 K, then In(

K_{1}

K_{2}

) is equal to :-

R = 8.314 J^{- 1} K^{- 1}

Q. Two reactions

R_{1}

and

R_{2}

have identical pre-exponential factors. Activation energy of

R_{1}

exceeds that of

R_{2}

10 k J m o l^{- 1}

. If

k_{1}

and

k_{2}

are rate constants for reactions

R_{1}

and

R_{2}

respectively at 300 K, then

l n (\frac{k_{2}}{k_{1}})

is equal to :
(R = 8.314

J m o l^{- 1} K^{- 1}

)

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For a consecutive reaction R1k1−→R2 R1k2−→R3 If initial concentration of R1 is 100 M and k1:k2=1:0.15 the value of [R2]max is [Given:k1=4.0×10−2min−1]

The correct option is A 71.55 M The values of [R2]max is given by the relation [R2]max=[R1]0[k2k1]k2(k2−k1) ∴[R2]max=100[0.15×10−21.0×10−2]0.015×10−2(1−0.15)×10−2=100[0.151]0.150.85=71.55M.

For a consecutive reaction $R_{1} k_{1} - \to R_{2} R_{1} k_{2} - \to R_{3}$ If initial concentration of $R_{1}$ is 100 M and $k_{1} : k_{2} = 1 : 0.15$ the value of $[R_{2}]_{m a x}$ is $[G i v e n : k_{1} = 4.0 \times 10^{- 2} m i n^{- 1}]$