Rate constant k - 2 - 103 mol-1 L s-1 and Ea - 2-0 - 102 kJ mol-1- What is the value of A when T - -

The correct option is B 2.0 × 10^3 #160; mol^ 1 L s^ 1 As we know, k = Ae^ E_a/RT If #160; T → ∞, k = A = 2 × 10^3 mol^ 1 L ...

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Standard XII

Chemistry

Arrhenius Equation

Rate constant...

Question

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 $T \to \infty$ ?

2.0 \times 10^{2} m o l^{- 1}

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2.0 \times 10^{3} m o l^{- 1} L s^{- 1}

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2.0 \times 10^{3} m o l L^{- 1} s^{- 1}

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2.0 \times 10^{3} m o l^{- 1} s^{- 1}

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Solution

The correct option is B $2.0 \times 10^{3} m o l^{- 1} L s^{- 1}$
As we know,

$k = A e^{- E_{a} / R T}$

If $T \to \infty,$

$k = A = 2 \times 10^{3} m o l^{- 1} L s^{- 1}$

Hence, option B is correct.

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Similar questions

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} . W h e n T \to \infty

A hypothetical reaction : $A_{(g)} + B_{(g)} ⇌ C_{(g)} + D_{(g)}$ occurs in a single step, the specific rate constant of forward reaction at TK is $2.0 \times 10^{- 3} m o l^{- 1} L s^{- 1}$ . When start is made with equimolar amounts of A and B, it is found that the concentration of A is twice that of C at equilibrium. The specific rate constant of the backward reaction is ?

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 $E_{a}$ ?

Q. For the reaction

2 A + B \to A_{2} B

rate

= k [A] [B]^{2}

with

k = 2.0 \times 10^{- 6} m o l^{- 2} L^{2} s^{- 1}

, what is the rate of the reaction when [A] is reduced to 0.060 molL^{-1} . the initial rate of reaction when

[A] = 0.1 m o l L^{- 1} a n d [B] = 0.2 m o l L^{- 1} i s 8 \times 10^{- 9} m o l L^{- 1} s^{- 1}

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 $E_{a}$ ?

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Rate constant k=2×103mol−1Ls−1andEa=2.0×102kJmol−1. What is the value of A when T→∞?

The correct option is B 2.0×103mol−1Ls−1As we know,k=Ae−Ea/RTIf T→∞,k=A=2×103mol−1Ls−1Hence, option B is correct.

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 $T \to \infty$ ?

The correct option is B $2.0 \times 10^{3} m o l^{- 1} L s^{- 1}$
As we know,

$k = A e^{- E_{a} / R T}$

If $T \to \infty,$

$k = A = 2 \times 10^{3} m o l^{- 1} L s^{- 1}$

Hence, option B is correct.