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Order of Reaction

Consider acti...

Question

Consider activation energy of two reaction is differ by $50 k J / m o l$ but they have equal expotential factor and same order of reaction. Then, calculate the log ratio of the rate constants of this reaction at $30^{\circ} C$ .
$(R = 8.314 J K^{- 1} m o l^{- 1})$

8.61

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4 \times 10^{5}

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14.3

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- 11.6

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Solution

The correct option is A $8.61$
We know that,
$l o g_{10} k = l o g_{10} A - \frac{E}{2.303 R T}$

So, $l o g_{10} k_{2} = l o g_{10} A - \frac{E_{2}}{2.303 R T} . . . . e q n (1)$

and $l o g_{10} k_{1} = l o g_{10} A - \frac{E_{1}}{2.303 R T} . . . . . e q n (2)$

Substracting equation (2) from (1), we get,

$l o g_{10} (\frac{k_{2}}{k_{1}}) = \frac{(E_{1} - E_{2})}{2.303 R T}$

$l o g_{10} (\frac{k_{2}}{k_{1}}) = \frac{50 \times 1000}{2.303 \times 8.314 \times 303}$

$l o g_{10} \frac{k_{2}}{k_{1}} = 8.61$
Hence, (a) is correct.

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Similar questions

Two reactions of the same order have equal exponential factors but their activation energies differ by $24.9 k J / m o l$ . Calculate the ratio of the rate constants of this reaction at $27^{\circ} C$ . $(R = 8.314 J K^{- 1} m o l^{- 1})$

Q. Two reactions of the same order have equal exponential factors but their activation energies differ by

24.9 k J / m o l

. Calculate the ratio between the rate constants of these reactions at

27^{o} C

(R = 8.31 J K^{- 1} {m o l}^{- 1})

Q. Two reactions of the same order have equal pre-exponential factors but their activation energies differ by

41.9 J / m o l

. Then the ratios between rate constants of these reactions at

600

K

is:

Q. Two reactions of same order have equal exponential factors but their activation energy differ by

24.9 k J m o l^{- 1}

. If the ratio between the rate constant of these reactions occurs at

27^{o} C

x \times 10^{4}

. Nearest integer value of

x

is_________.(R=8.314

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

)

Q. Two reactions

A

and

B

of the same order have equal pre-exponential factors but activation energy for

A

is greater than that of

B

24.9 k J / m o l

. Calculate the ratio(

\frac{k_{B}}{k_{A}}

) at

27^{\circ} C .

Here,

k_{A}

and

k_{B}

are rate constants for reactions

A

and

B

, respectively.

(R = 8.31 J K^{- 1} m o l^{- 1})

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Consider activation energy of two reaction is differ by 50kJ/mol but they have equal expotential factor and same order of reaction. Then, calculate the log ratio of the rate constants of this reaction at 30∘C. (R=8.314J K−1mol−1)

The correct option is A 8.61We know that, log10k=log10A−E2.303RT So, log10k2=log10A−E22.303RT....eqn(1) and log10k1=log10A−E12.303RT.....eqn(2) Substracting equation (2) from (1), we get, log10(k2k1)=(E1−E2)2.303RT log10(k2k1)=50×10002.303×8.314×303 log10 k2k1=8.61 Hence, (a) is correct.

Consider activation energy of two reaction is differ by $50 k J / m o l$ but they have equal expotential factor and same order of reaction. Then, calculate the log ratio of the rate constants of this reaction at $30^{\circ} C$ .
$(R = 8.314 J K^{- 1} m o l^{- 1})$