At 407 K- the rate constant of a chemical reaction is 9-5-10-5s-1 and at 420 K- the rate constant is 1-9-10-4s-1- Find the value of logA where A is the frequency factor of the reaction- Take-log 9-5 - 0-97

Expression for rate constant at two different temperature is given by, nbsp; nbsp;log_10k_2k_1=E_a2.303× R[T_2 T_1T_1T_2] Given: nbsp; k_1=9.5×10^ 5s^ 1 ...

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Standard VII

Chemistry

Half Life

At 407 K, the...

Question

At $407 K$ , the rate constant of a chemical reaction is $9.5 \times 10^{- 5} s^{- 1}$ and at $420 K$ , the rate constant is $1.9 \times 10^{- 4} s^{- 1}$ . Find the value of $l o g (A)$ where A is the frequency factor of the reaction.
Take:
$l o g (9.5) = 0.97$

5.68

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Solution

The correct option is A $5.68$
Expression for rate constant at two different temperature is given by,
$l o g_{10} \frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303 \times R} [\frac{T_{2} - T_{1}}{T_{1} T_{2}}]$
Given: $k_{1} = 9.5 \times 10^{- 5} s^{- 1}; k_{2} = 1.9 \times 10^{- 4} s^{- 1};$
$R = 8.314 J m o l^{- 1} K^{- 1};$
$T_{1} = 407 K$ and $T_{2} = 420 K$
Substituting the values in above equation
$l o g_{10} \frac{1.9 \times 10^{- 4}}{9.5 \times 10^{- 5}} = \frac{E_{a}}{2.303 \times 8.314} [\frac{420 - 407}{420 \times 407}]$

$E_{a} = 75531.05 J m o l^{- 1}$
Arrhenius equation of a reaction at $407 K$
$k_{1} = A e^{- E_{a} / R T_{1}}$
Taking log on both sides,
$l o g k_{1} = l o g A - \frac{E_{a}}{2.303 R T_{1}}$

Substituting the values,
$l o g (9.5 \times 10^{- 5}) = l o g A - \frac{75531}{2.303 \times 8.314 \times 407}$

$\Rightarrow - 4.02 = l o g A - 9.7$

$l o g A = 5.68$
Hence, (a) is correct.

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At 407 K, the rate constant of a chemical reaction is 9.5×10−5s−1 and at 420 K, the rate constant is 1.9×10−4s−1. Find the value of log(A) where A is the frequency factor of the reaction. Take: log (9.5)=0.97

At $407 K$ , the rate constant of a chemical reaction is $9.5 \times 10^{- 5} s^{- 1}$ and at $420 K$ , the rate constant is $1.9 \times 10^{- 4} s^{- 1}$ . Find the value of $l o g (A)$ where A is the frequency factor of the reaction.
Take:
$l o g (9.5) = 0.97$