In the Arrhenius equation for a certain reaction the value of A and Ea (activation energy) are 4 × 1013 sec-1 and 98.6 kJ mol-1 respectively. If the reaction is of first order, the temperature at which its half-life period is 10 minutes is

Question

In the Arrhenius equation for a certain reaction the value of A and Ea (activation energy) are 4 × 1013 sec-1 and 98.6 kJ mol-1 respectively. If the reaction is of first order, the temperature at which its half-life period is 10 minutes is (A) 280 K (B) 290 K (C) 311.35 K (D) 418.26 K

Tardigrade Admin · Accepted Answer

Calculation of K We know that k=(0.693/t12)=(0.693/10×60)(t1/2=10×60 sec)=1.1558×10-3 log k=log A-(Ea/2.303 RT) log 1.155×10-3=log4×1013-(98.6/2.303×8.314×10-3× T) On usual calculation, T = 311.3 5 K

Q. In the Arrhenius equation for a certain reaction the value of $A$ and $E_{a}$ (activation energy) are $4 \times 1 0^{13} se c^{- 1}$ and $98.6 k J m o l^{- 1}$ respectively. If the reaction is of first order, the temperature at which its half-life period is 10 minutes is

$280 K$

$290 K$

$311.35 K$

$418.26 K$

Q. In the Arrhenius equation for a certain reaction the value of A and Ea​ (activation energy) are 4×1013sec−1 and 98.6kJmol−1 respectively. If the reaction is of first order, the temperature at which its half-life period is 10 minutes is

280K

290K

311.35K

418.26K

Calculation of K We know that k=t120.693​=10×600.693​(t1/2​=10×60sec)=1.1558×10−3 log k=logA−2.303RTEa​​ log 1.155×10−3=log4×1013−2.303×8.314×10−3×T98.6​ On usual calculation, T=311.35K

Q. In the Arrhenius equation for a certain reaction the value of $A$ and $E_{a}$ (activation energy) are $4 \times 1 0^{13} se c^{- 1}$ and $98.6 k J m o l^{- 1}$ respectively. If the reaction is of first order, the temperature at which its half-life period is 10 minutes is

$280 K$

$290 K$

$311.35 K$

$418.26 K$