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Right Hand Limit

A⟶ B, K A =...

Question

$A ⟶ B$ , $K_{A} = 10^{15} e^{- 2000 / T}$
$C ⟶ D$ , $K_{C} = 10^{14} e^{- 1000 / T}$
$T e m p e r a t u r e T; K a t w h i c h (K_{A} = K_{C})$ ?

A

1000 K

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B

2000 K

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C

\frac{2000}{2.303} K

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D

\frac{1000}{2.303} K

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Solution

The correct option is C $\frac{1000}{2.303} K$
As given,
$A ⟶ B; K_{A} = 10^{15} e^{- 2000 / T}$
$C ⟶ D; K_{C} = 10^{14} e^{- 1000 / T}$

So, for $K_{A} = K_{C}$
$\Rightarrow 10^{15} e^{- 2000 / T} = 10^{14} e^{- 1000 / T}$
$\Rightarrow 10 = e^{1000 / T}$

Taking natural logarithm on both sides:-
$\Rightarrow l n 10 = \frac{1000}{T}$
$\Rightarrow T = \frac{1000}{l n 10} K = \frac{1000}{2.303 {log}_{10}^{10}} = \frac{1000}{2.303} K$

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Similar questions

A \to B; K_{A} = 10^{15} e^{\frac{- 2000}{T}}

C \to D; K_{C} = 10^{14} e^{\frac{- 1000}{T}}

Temperature T in terms of K at which

K_{A} = K_{C}

Q. The rate constants

K_{1}

K_{2}

for two different reactions are

10^{15} e^{- \frac{1000}{T}}

10^{16} e^{- \frac{2000}{T}}

respectively. The temperature at which

K_{1} = K_{2}

Q. The rate constants

k_{1}

k_{2}

for two different reactions are

10^{16} . e^{- 2000 / T}

10^{15} . e^{- 1000 / T}

respectively, the temperature at which

k_{1} = k_{2}

A \to B K_{A} = 10^{15} e^{2000 / T} C \to D K_{C} = 10^{14} e^{1000 / T}

T (K)

(K_{A} = K_{C})

A \to B K_{A} = 10^{15} e^{2000 / T}

C \to D K_{C} = 10^{14} e^{1000 / T}

Temperature T K at which

(K_{A} = K_{C})

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