For the reaction- -CrH2O6-3-SCN- -Cr9H2O5NCS-2-H2OThe rate law is - r - k-CrH2O63-SCN- The value of k is 2-0 - 10-6 L mol-1 s-1 at 140 C and 2-2 - 10-5 L mol-1 s-1 - at 300 C What is the value of Ea-

The correct option is A 26 kcal mol^ 1 Using Arrhenius equation #160; T_1 = 273 + 14 = 287 K , #160; T_2 = 273 + 30 = 303 K ...

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Byju's Answer

Standard XII

Chemistry

Arrhenius Equation

For the react...

Question

For the reaction:
$[C r (H_{2} O)_{6}]^{3 +} + [S C N^{⊖}] \to [C r 9 H_{2} O)_{5} N C S]^{2 +} H_{2} O$
The rate law is : $r = k [C r (H_{2} O)_{6}^{3 +} [S C N^{⊖}] .$
The value of k is $2.0 \times 10^{- 6} L m o l^{- 1} s^{- 1}$ at $14^{0}$ C and $2.2 \times 10^{- 5} L m o l^{- 1} s^{- 1}$ , at $30^{0}$ C
What is the value of $E_{a}$ ?

26 k c a l

m o l^{- 1}

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2.6 k c a l

m o l^{- 1}

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2600 k c a l

m o l^{- 1}

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260 k c a l

m o l^{- 1}

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Solution

The correct option is A $26 k c a l$ $m o l^{- 1}$
Using Arrhenius equation
$T_{1}$ $= 273 + 14 = 287 K$ ,
$T_{2}$ = $273 + 30 = 303 K$
$l o g \frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303 R} [\frac{T_{2} - T_{1}}{T_{1} \times T_{2}}]$
$l o g \frac{2.2 \times 10^{- 5}}{2.0 \times 10^{- 6}} = \frac{E_{a}}{2.303 \times 2 \times 10^{- 3}} (\frac{16}{287 \times 303})$
$E_{a} = \frac{l o g (11) \times 2.303 \times 2 \times 10^{- 3} \times 287303}{16}$
$= \frac{1.0414 \times 2.303 \times 2 \times 287 \times 303}{16000}$
$= 26 k c a l m o l^{- 1}$

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Similar questions

For the reaction:

$[C r (H_{2} O)_{6}]^{3 +} + [S C N^{⊖}] \to [C r 9 H_{2} O)_{5} N C S]^{2 +} H_{2} O$

The rate law is : $r = k [C r (H_{2} O)_{6}^{3 +} [S C N^{⊖}] .$

The value of k is $2.0 \times 10^{- 6} L m o l^{- 1} s^{- 1}$ at 14 $^{0}$ C and $2.2 \times 10^{- 5} L m o l^{- 1} s^{- 1} a t 30^{\circ} C .$ What is the value of $E_{a}$ ?

Q. For the reaction 2A + B + C

\to A_{2} B

+ C the rate law is found to be Rate =k[A][B]

^{2}

with

k = 2 \times 10^{- 6} m o l^{- 2} L^{2} s^{- 1}

The initial rate of reaction with

[A] = 0.1 m o l L^{- 1}, [B] = 0.2 m o l L^{- 1} a n d [C] = 0.8 m o l L^{- 1}

Q. Rate constant

k = 2 \times 10^{3} m o l^{- 1} L s^{- 1} a n d E_{a} = 2.0 \times 10^{2} k J m o l^{- 1}

. What is the value of A when

T \to \infty

Q. Identify the order of the reaction from the following rate constants respectively
a.

k = 7.06 \times 10^{- 3} {mol}^{- 1} {L s}^{- 1}

k = 5.60 \times 10^{- 5} s^{- 1}

k = 1.25 \times 10^{- 2} {mol L}^{- 1} s^{- 1}

k = 1.25 \times 10^{- 2} {mol}^{- 2} L^{2} s^{- 1}

Q. Following data are given for the reaction between

A

and

B

$[A] / m o l L^{- 1}$	$[B] / m o l L^{- 1}$	Initial rate/ $m o l L^{- 1} s^{- 1}$ at $300 K$	$320 K$
$2.5 \times 10^{- 4}$	$3.0 \times 10^{- 5}$	$5.0 \times 10^{- 4}$	$2.0 \times 10^{- 3}$
$5.0 \times 10^{- 4}$	$6.0 \times 10^{- 5}$	$4.0 \times 10^{- 3}$	-
$1.0 \times 10^{- 3}$	$6.0 \times 10^{- 5}$	$1.6 \times 10^{- 2}$	-

The rate constant at

300 K

is:

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For the reaction:[Cr(H2O)6]3++[SCN⊖]→[Cr9H2O)5NCS]2+H2OThe rate law is : r=k[Cr(H2O)3+6[SCN⊖]. The value of k is 2.0×10−6Lmol−1s−1 at 140 C and 2.2×10−5Lmol−1s−1 , at 300 C What is the value of Ea?

The correct option is A 26kcal mol−1Using Arrhenius equation T1 =273+14=287K, T2 = 273+30=303Klogk2k1=Ea2.303R[T2−T1T1×T2]log2.2×10−52.0×10−6=Ea2.303×2×10−3(16287×303)Ea=log(11)×2.303×2×10−3×28730316=1.0414×2.303×2×287×30316000 =26kcalmol−1