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Chemistry Question on Chemical Kinetics

The rate constant for the first order decomposition of H2O2H_2O _2 is given by the following equation:
log k=14.34−1.25×104 K/Tlog\ k = 14.34 - 1.25 \times 10^4\ K/T
Calculate EaE_a for this reaction and at what temperature will its half-period be 256 minutes?

Answer

Arrhenius equation given by,Arrhenius \ equation \ given\ by,

k=Ae−EaRTk = Ae^{\frac {-E_a}{RT}}

ln k=ln A−EaRTln\ k = ln \ A - \frac {E_a}{RT}

ln k=log A−EaRTln\ k = log \ A - \frac {E_a}{RT}

⇒log k=log A−Ea2.303 Tlog\ k = log \ A - \frac {E_a}{2.303\ T} .....(i)

The given equation is

log k=14.34−1.25×104 K/Tlog\ k = 14.34-1.25×10^4\ K/T ......(ii)

From equation (i) and (ii), we obtain

Ea2.303 RT=1.25×104 K/T\frac {E_a}{2.303 \ RT} = 1.25×10^4\ K/T

⇒ Ea=1.25×104K×2.303×REa=1.25×10^4K×2.303×R

= 1.25×104K×2.303×8.314JK−1mol−11.25 × 10^4K × 2.303 × 8.314 J K^{- 1}mol^ {- 1}

= 239339.3 Jmol−1(approximately)239339.3 \ J mol^{-1} (approximately)

= 239.34 kJmol−1239.34\ kJ mol^{-1}

Also, when t12=256 minutest_{\frac 12}=256 \ minutes,

k=0.693t1/2k = \frac {0.693}{t_{1/2}}

k=0.693256k = \frac {0.693}{256}

k=2.707×10−3min−1k= 2.707×10^{-3} min^{-1}

k=4.51×10−5s−1k= 4.51 × 10^{-5} s^{-1}

It is also given that, log k=14.34−1.25×104K/Tlog\ k= 14.34 - 1.25 × 10^4 K/T

⇒log(4.51×10−5)=14.34−1.25×104K/Tlog(4.51×10^{-5})=14.34-1.25×10^4 K/T

log (0.654−05)=14.34−1.25×104K/Tlog \ (0.654-05)=14.34-1.25×10^4 K/T

1.25×104KT=18.686\frac {1.25×10^4K}{T} = 18.686

⇒T=1.25×104 K18.686T =\frac {1.25 \times 10^4 \ K}{18.686}

T=668.95 KT = 668.95 \ K

T=669 K (approximately)T = 669\ K\ (approximately)