For a reaction, A longrightarrow B ; if log 10 K( sec -1)=14-(1.25 × 104/T) K, the frequency factor and energy of activation for the reaction are

Question

For a reaction, A longrightarrow B ; if log 10 K( sec -1)=14-(1.25 × 104/T) K, the frequency factor and energy of activation for the reaction are (A) 1014 sec -1, 239.34 kJ (B) 14,57.6 kcal (C) 1014 sec -1, 23.93 kJ (D) 1014 sec , 5.76 kcal

Tardigrade Admin · Accepted Answer

log 10 K( sec -1)=14-(1.25 × 104/T) log 10 K= log 10 A-(Ea/2.303 R T) log 10 A=14 A=1014 sec -1 and (Ea/2.303 R)=1.25 × 104 Ea=1.25 × 2.303 × 104 × 8.314 J ≈ 239 × 103 J =239 kJ

Q. For a reaction, $A ⟶ B;$ if $lo g_{10} K (se c^{- 1}) = 14 - \frac{1.25 \times 1 0 ^{4}}{T} K$ , the frequency factor and energy of activation for the reaction are

$1 0^{14} se c^{- 1}, 239.34 k J$

$14, 57.6 k c a l$

$1 0^{14} se c^{- 1}, 23.93 k J$

$1 0^{14} sec, 5.76 k c a l$

Q. For a reaction, A⟶B; if log10​K(sec−1)=14−T1.25×104​K, the frequency factor and energy of activation for the reaction are

1014sec−1,239.34kJ

14,57.6kcal

1014sec−1,23.93kJ

1014sec,5.76kcal

log10​K(sec−1)=14−T1.25×104​ log10​K=log10​A−2.303RTEa​​ log10​A=14 A=1014sec−1 and 2.303REa​​=1.25×104 Ea​=1.25×2.303×104×8.314J≈239×103J=239kJ

Q. For a reaction, $A ⟶ B;$ if $lo g_{10} K (se c^{- 1}) = 14 - \frac{1.25 \times 1 0 ^{4}}{T} K$ , the frequency factor and energy of activation for the reaction are

$1 0^{14} se c^{- 1}, 239.34 k J$

$14, 57.6 k c a l$

$1 0^{14} se c^{- 1}, 23.93 k J$

$1 0^{14} sec, 5.76 k c a l$