For the reaction A - B- the rate constant ks-1 is given by log10 k-20-35-2-47- 103TThe energy of activation in kJ mol-1 is Nearest integer-Given - R-8-314 JK-1mol-1-

Log_10k=20.35 (2.47× 10^3)T ....(1) log k=log A E_a2.303 RT ....(ii) Comparing (i) and (ii), E_a2.303 RT=2.47× 10^3T E_a=2.47× 10^3× 2.303× 8.314 E_a= 47 ...

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Standard XII

Business Studies

Significance of Management Principles

For the react...

Question

For the reaction $A \to B$ , the rate constant $k (s^{- 1})$ is given by

${log}_{10} k = 20.35 - \frac{(2.47 \times 10^{3})}{T}$

The energy of activation in ${kJ mol}^{- 1}$ is
(Nearest integer)

[Given : R=8.314 ${JK}^{- 1} {mol}^{- 1}$ ]

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47.00

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

47.0

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Solution

${log}_{10} k = 20.35 - \frac{(2.47 \times 10^{3})}{T} . . . . (1)$

$log k = log A - \frac{E_{a}}{2.303 R T} . . . . (i i)$

Comparing (i) and (ii),

$\frac{E_{a}}{2.303 R T} = \frac{2.47 \times 10^{3}}{T}$

$E_{a} = 2.47 \times 10^{3} \times 2.303 \times 8.314$

$E_{a} = 47293.44 {J mol}^{- 1}$
$E_{a} = 47.2934 {kJ mol}^{- 1}$
Nearest integer is 47

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For the reaction A→B, the rate constant k(s−1) is given by log10 k=20.35−(2.47×103)T The energy of activation in kJ mol−1 is (Nearest integer) [Given : R=8.314 JK−1mol−1]

log10k=20.35−(2.47×103)T ....(1) logk=logA−Ea2.303 RT ....(ii) Comparing (i) and (ii), Ea2.303 RT=2.47×103T Ea=2.47×103×2.303×8.314 Ea=47293.44J mol−1 Ea=47.2934 kJ mol−1 Nearest integer is 47

For the reaction $A \to B$ , the rate constant $k (s^{- 1})$ is given by

${log}_{10} k = 20.35 - \frac{(2.47 \times 10^{3})}{T}$

The energy of activation in ${kJ mol}^{- 1}$ is
(Nearest integer)

[Given : R=8.314 ${JK}^{- 1} {mol}^{- 1}$ ]