Derive Arrhenius equation from Van’t Hoff Equation. | Chemical Kinetics | Physical Chemistry

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The van't Hoff equation relates the change in the equilibrium constant left( { K }_{ eq } right) of a chemical reaction to the change in temperature left( T right) . This Van't Hoff's relation is: boxed { frac { dleft( ln { K } right) }{ dT } = frac { triangle E }{ R{ T }^{ 2 } } } ... (1) where triangle E is energy of activation, R is gas constant, T is temperature and K is the equilibrium constant for the reaction. Consider a reversible reaction, A + B begin{matrix} { k }_{ 1 } Longleftrightarrow { k }_{ 2 } end{matrix} C + D Equilibrium constant K can be taken as frac { { k }_{ 1 } }{ { k }_{ 2 } } , where { k }_{ 1 } and { k }_{ 2 } are the velocity constants for the forward and back ward reactions. If { E }_{ 1 } and { E }_{ 2 } are the activation energies of the reactant and the product, then substituting these values in the above Van't Hoff's equation we get, frac { dleft( ln { frac { { k }_{ 1 } }{ { k }_{ 2 } } } right) }{ dT } = frac { left( { E }_{ 1 } - { E }_{ 2 } right) }{ R{ T }^{ 2 } } frac { dleft( ln { { k }_{ 1 } } - ln { { k }_{ 2 } } right) }{ dT } = frac { left( { E }_{ 1 } - { E }_{ 2 } right) }{ R{ T }^{ 2 } } frac { dleft( ln { { k }_{ 1 } } right) }{ dT } - frac { dleft( ln { { k }_{ 2 } } right) }{ dT } = frac { { E }_{ 1 } }{ R{ T }^{ 2 } } - frac { { E }_{ 2 } }{ R{ T }^{ 2 } } ... (2) This equation can be written in two parts as follows: frac { dleft( ln { { k }_{ 1 } } right) }{ dT } = frac { { E }_{ 1 } }{ R{ T }^{ 2 } } ... (3) frac { dleft( ln { { k }_{ 2 } } right) }{ dT } = frac { { E }_{ 2 } }{ R{ T }^{ 2 } } ... (4) or in general, frac { dleft( ln { k } right) }{ dT } = frac { E }{ R{ T }^{ 2 } } ... (5) on rearranging we can write, dleft( ln { k } right) = frac { E }{ R } frac { dT }{ { T }^{ 2 } } This expression on integration gives, int { dleft( ln { k } right) } = frac { E }{ R } int { frac { 1 }{ { T }^{ 2 } } dT } ln { k } = frac { E }{ R } left( frac { -1 }{ T } right) + I where I is the constant of integration ln { k } = I + left( -frac { E }{ RT } right) on taking antilog of exponential we get, k = { e }^{ left[ I + left( -frac { E }{ RT } right) right] } k = { e }^{ left[ I right] } times { e }^{ left( sfrac { -E }{ RT } right) } boxed { k = A{ e }^{ sfrac { -E }{ RT } } } ... (7) Where { e }^{ left[ I right] } = A is another constant known as frequency factor. Equation (7) is Arrhenius equation. Arrhenius gave an equation to study the effect of temperature on rate of reaction.
Numerical: A first order reaction was found to have an energy of activation 1.25 times { 10 }^{ 5 }J/mol. Calculate the temperature at which the reaction will have a half-life of one minute. The frequency factor A in the Arrhenius quation is 5 times { 10 }^{ 13 }{ s }^{ -1 }. (R = 8.314 J{ K }^{ -1 }{ mol }^{ -1 }) Solution: We have, Energy of Activation E = 1.25 times { 10 }^{ 5 }J{ mol }^{ -1 } The half-life period left( { t }_{ sfrac { 1 }{ 2 } } right) = 1 min = 60 Sec therefore k = frac { 0.693 }{ { t }_{ sfrac { 1 }{ 2 } } } = frac { 0.693 }{ 60 } = 0.01155{ s }^{ -1 } Frequency factor A = 5 times { 10 }^{ 13 }{ s }^{ -1 } Temperature T = ? According to Arrheniuse quation, boxed { k = A.{ e }^{ left( sfrac { -E }{ RT } right) } } Substituting the values, left( 0.01155{ s }^{ -1 } right) = left( 5 times { 10 }^{ 13 }{ s }^{ -1 } right) times { e }^{ left( sfrac { -E }{ RT } right) } therefore { e }^{ left( sfrac { -E }{ RT } right) } = frac { 0.01155 }{ 5 times { 10 }^{ 13 } } = 2.31 times { 10 }^{ -16 } On taking natural log frac { -E }{ RT } = ln { left( 2.31 times { 10 }^{ -16 } right) } = -36.004 frac { -left( 1.25 times { 10 }^{ 5 } right) }{ 8.314 times T } = -36.004 therefore T = frac { 1.25 times { 10 }^{ 5 } }{ 8.314 times 36.004 } = 417.59K

[@priyaparamitapatra4756]

Very clear and very nice explanation.... Thank u sir..... 🙏

[@priyabiswal9629]

Sir thanks many videos watched but couldn't understand how to take the exponent properly now I got clear.

[@roshanthomas2761]

Very clear explanation! Thank you.

[@Sarah-hq3mw]

How do you get the first equation? When I'm looking at Van't Hoff's equation it's usually delta H, how does it become the Activation Energy?

[@umangagarwal8726]

True. Same doubt. Did you clear it?

[@ahammedhassan2176]

thank you very much

[@successfulsuccess9227]

Am happy , it's really helpful thanks alot

[@Edmerls]

Glad it was helpful!

[@ujjwal124]

Searching for it

[@SamirPanigrahy-vi3bd]

Ty sir

[@ishangor3496]

Very nice

[@Dumela_go_Jesu_Kriste]

is 0.693 the k value for this specific reaction, or does it have to somehow be calculated from somewhere?

[@shakshisekar2169]

Hi! This is the formula for half life time period for a first-order reaction

[@saurabhnagarkoti9588]

It is nothing but the value of ln2

[@pritikalra3334]

Yes