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sladkih [1.3K]
3 years ago
5

He rate constant of a reaction is 4.55 × 10−5 l/mol·s at 195°c and 8.75 × 10−3 l/mol·s at 258°c. what is the activation energy o

f the reaction? enter your answer in scientific notation.
Chemistry
2 answers:
Xelga [282]3 years ago
8 0

Answer : The activation energy of the reaction is, 17.285\times 10^4kJ/mole

Solution :  

The relation between the rate constant the activation energy is,  

\log \frac{K_2}{K_1}=\frac{Ea}{2.303\times R}\times [\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_1 = initial rate constant = 4.55\times 10^{-5}L/mole\text{ s}

K_2 = final rate constant = 8.75\times 10^{-3}L/mole\text{ s}

T_1 = initial temperature = 195^oC=273+195=468K

T_2 = final temperature = 258^oC=273+258=531K

R = gas constant = 8.314 kJ/moleK

Ea = activation energy

Now put all the given values in the above formula, we get the activation energy.

\log \frac{8.75\times 10^{-3}L/mole\text{ s}}{4.55\times 10^{-5}L/mole\text{ s}}=\frac{Ea}{2.303\times (8.314kJ/moleK)}\times [\frac{1}{468K}-\frac{1}{531K}]

Ea=17.285\times 10^4kJ/mole

Therefore, the activation energy of the reaction is, 17.285\times 10^4kJ/mole

garik1379 [7]3 years ago
8 0
<h3><u>Answer;</u></h3>

139.8 kJ/mol

<h3><u>Explanation</u>;</h3>

The activation energy is given by the formula;

Eₐ = R•ln[k₂/k₁]•T₁T₂/(T₂ – T₁)  

After converting ˚C to K:

T1 = 531 K

T2 = 468 k (after converting ˚C to K):  

R = 8.314 J/mol•K

Eₐ = (8.314 J/mol•K) × ln[(3.2 × 10⁻³)/(4.5 × 10⁻⁵)] × (531 K)(468 K)/(531 – 468 K)  

Eₐ = 139.8 kJ/mol

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