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Arrhenius Equation

A → B KA ...

Question

$A \to B K_{A} = 10^{15} e^{2000 / T} C \to D K_{C} = 10^{14} e^{1000 / T}$
Temperature $T (K)$ at which $(K_{A} = K_{C})$ is:

1000 K

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2000 K

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(2000 / 2.303) K

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(1000 / 2.303) K

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Solution

The correct option is C $(1000 / 2.303) K$
$A \to B K_{A} = 10^{15} e^{2000 / T} C \to D K_{C} = 10^{14} e^{1000 / T} A t T (K_{A} = K_{C}) C_{0}^{15} e^{2000 / T} = 10^{14} e^{1000 / T} 10 e^{2000 / T} = e^{1000 / T} 10 = e^{1000 / T} l o g_{e} 10 = 1000 / T T = \frac{1000}{l o g_{e} 10} T = (\frac{1000}{2.303}) k$

Suggest Corrections

Similar questions

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

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

(K_{A} = K_{C})

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

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

K_{A} = K_{C}

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})

K_{1}

K_{2}

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

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

K_{1} = K_{2}

T_{1} = 1000 K

1.4 \times 10^{- 6} m

2.8 \times 10^{- 6} m

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