The rate constant of a reaction is 1.5 × 10-3 at 25° C and 2.1 × 10-2 at 60° C. The activation energy is

Question

The rate constant of a reaction is 1.5 × 10-3 at 25° C and 2.1 × 10-2 at 60° C. The activation energy is (A) (35/333) R log e (2.1 × 10-2/1.5 × 10-2) (B) (298 × 333/35) R log e (21/1.5) (C) (298 × 333/35) R log e 2.1 (D) (298 × 333/35) R log e (2.1/1.5)

Tardigrade Admin · Accepted Answer

T1=273+25=298 K T2=273+60=333 K log (k2/k1)=(Ea/2.3 R)((T2-T1/T1 T2)) log e (k2/k1)=(Ea/R)((T2-T1/T1 T2)) or log e (2.1 × 10-2/1.5 × 10-3)=(Ea/R)((35/333 × 298)) ∴ Ea=(298 × 333/35) × R × log e (21/1.5)

Q. The rate constant of a reaction is $1.5 \times 1 0^{- 3}$ at $2 5^{\circ} C$ and $2.1 \times 1 0^{- 2}$ at $6 0^{\circ} C$ . The activation energy is

$\frac{35}{333} R lo g_{e} \frac{2.1 \times 1 0 ^{- 2}}{1.5 \times 1 0 ^{- 2}}$

$\frac{298 \times 333}{35} R lo g_{e} \frac{21}{1.5}$

$\frac{298 \times 333}{35} R lo g_{e} 2.1$

$\frac{298 \times 333}{35} R lo g_{e} \frac{2.1}{1.5}$

Q. The rate constant of a reaction is 1.5×10−3 at 25∘C and 2.1×10−2 at 60∘C. The activation energy is

33335​Rloge​1.5×10−22.1×10−2​

35298×333​Rloge​1.521​

35298×333​Rloge​2.1

35298×333​Rloge​1.52.1​

T1​=273+25=298K T2​=273+60=333K logk1​k2​​=2.3REa​​(T1​T2​T2​−T1​​) loge​k1​k2​​=REa​​(T1​T2​T2​−T1​​) or loge​1.5×10−32.1×10−2​=REa​​(333×29835​) ∴Ea​=35298×333​×R×loge​1.521​

Q. The rate constant of a reaction is $1.5 \times 1 0^{- 3}$ at $2 5^{\circ} C$ and $2.1 \times 1 0^{- 2}$ at $6 0^{\circ} C$ . The activation energy is

$\frac{35}{333} R lo g_{e} \frac{2.1 \times 1 0 ^{- 2}}{1.5 \times 1 0 ^{- 2}}$

$\frac{298 \times 333}{35} R lo g_{e} \frac{21}{1.5}$

$\frac{298 \times 333}{35} R lo g_{e} 2.1$

$\frac{298 \times 333}{35} R lo g_{e} \frac{2.1}{1.5}$