The time required for 10 % completion of a first order reaction at 298 K is equal to that required for its 25 % completion at 308 . If the value of A is 4 × 1010 s -1 . The value of Ea is ( log 2.728=0.436)

Question

The time required for 10 % completion of a first order reaction at 298 K is equal to that required for its 25 % completion at 308 . If the value of A is 4 × 1010 s -1 . The value of Ea is ( log 2.728=0.436) (A) 76.623 kJ / mol (B) 100 kJ / mol (C) 94 kJ / mol (D) 52.261 kJ / mol

Tardigrade Admin · Accepted Answer

Apply, k t=2.303 log ((A)0/(A)) k 1 × t =2.303 log (( A )0/90( A )0) × 100...(1) k 2 × t =2.303 log (( A )0/75( A )0) × 100...(2) (k2/k1)=2.728 Now, log ( k 2/ k 1)=( E a /2.303 R ) (( T 2- T 1)/ T 1 T 2) log (2.78)=( E a /2.303 × 8.314 × 10-3)((308 -298/308 × 298)) Ea=76.623 kJ / mol

Q. The time required for $10%$ completion of a first order reaction at $298 K$ is equal to that required for its $25%$ completion at $308$ . If the value of $A$ is $4 \times 1 0^{10} s^{- 1} .$ The value of $E_{a}$ is $(lo g 2.728 = 0.436)$

$76.623 k J / m o l$

$100 k J / m o l$

$94 k J / m o l$

$52.261 k J / m o l$

Q. The time required for 10% completion of a first order reaction at 298K is equal to that required for its 25% completion at 308 . If the value of A is 4×1010s−1. The value of Ea​ is (log2.728=0.436)

76.623kJ/mol

100kJ/mol

94kJ/mol

52.261kJ/mol

Apply, kt=2.303log(A)(A)0​​ k1​×t=2.303log90(A)0​(A)0​​×100...(1) k2​×t=2.303log75(A)0​(A)0​​×100...(2) k1​k2​​=2.728 Now, logk1​k2​​=2.303REa​​T1​T2​(T2​−T1​)​ log(2.78)=2.303×8.314×10−3Ea​​(308×298308−298​) Ea​=76.623kJ/mol

Q. The time required for $10%$ completion of a first order reaction at $298 K$ is equal to that required for its $25%$ completion at $308$ . If the value of $A$ is $4 \times 1 0^{10} s^{- 1} .$ The value of $E_{a}$ is $(lo g 2.728 = 0.436)$

$76.623 k J / m o l$

$100 k J / m o l$

$94 k J / m o l$

$52.261 k J / m o l$