You visited us 1 times! Enjoying our articles? Unlock Full Access!

Arrhenius Equation

Rate constant...

Question

Rate constant of a reaction changes by $2$ % by ${0.1}^{o} C$ rise in temperature at $25^{o} C$ . The standard heat of reaction is $121.6$ $k J$ ${m o l}^{- 1}$ . Calculate $E_{a}$ of reverse reaction.

Open in App

Solution

$log (\frac{k_{2}}{k_{1}}) = \frac{E_{a}}{2.303 R} (\frac{1}{T_{1}} - \frac{1}{T_{2}})$
$log \frac{102}{100} = \frac{E_{a}}{2.303 \times 8.314} (\frac{1}{298} - \frac{1}{298.1})$
$E_{a} = 1.463 \times 10^{5} J / m o l = 146.3 k J / m o l$
$Δ H = E_{f} - E_{b}$
$121.6 = 146.3 - E_{b}$
$E_{b} = 24.7 k J / m o l$

Suggest Corrections

Similar questions

10.6

25^{o} C

35^{o} C

k = A e^{- E_{a} / R T}

25^{o} C

162

k J

350

k J

O H^{-}

C H_{3} C l (a q) + O H (a q)_{f} K_{r} C H_{3} O H (a q) + C l (a q)

25^{o} C

6 \times 10^{6} M^{1} s^{- 1}

K_{c}

1 \times 10^{16}

25^{o} C

k = A e^{- E_{a} / R T}

l o g_{10} \frac{k_{c a}}{k_{u n}}

27^{\circ} C

k_{c a}

k_{u n}

162 k J m o l^{- 1}

350 k J m o l^{- 1}

E_{a}

m o l^{- 1}

m o l^{- 1}

Join BYJU'S Learning Program