Question

consider the following reactions reaction 1: a gives b;ea=28.31kj reaction 2: c gives d;ea=20 kj if the initial temperature is 400 k for both reaction and the rate constants are equal then find the ratio of arrhenius factor a1:a2 for reaction 1 and 2​

Answer 1

Answer:

Solution

The activation energy can be determined using the equation:

ln(k2/k1) = Ea/R x (1/T1 - 1/T2)

where

Ea = the activation energy of the reaction in J/mol

R = the ideal gas constant = 8.3145 J/K·mol

T1 and T2 = absolute temperatures (in Kelvin)

k1 and k2 = the reaction rate constants at T1 and T2

Answer 2

Final answer:

Ratio of Arrhenius factor $A_{1}:A_{2}$ for reaction 1 and 2 $=12.1679:1$

Given that: We are given, reaction:

reaction 1:  $A$ → $B$        $E_{a_{1}} = 28.31 \frac{kJ}{mol}$

reaction 2: $C$ → $D$       $E_{a_{2}} = 20 \frac{kJ}{mol}$  

Temperature both reaction $T= 400$ $K$      

To find: We have to find the ratio of Arrhenius factor $A_{1}:A_{2}$ for reaction 1 and 2​.

Explanation:

• The Arrhenius Equation is: $k=Ae^{\frac{-E_{a}}{RT}}$

Where,

T= Kelvin temperature

k = Rate constant of the reaction

A = Frequency factor or Arrhenius factor

$E_{a}$ = Activation energy of the reaction

R = Universal gas constant = $8.314JK^{-1}mol^{-1}$

• Given reactions:

1)  $A$ → $B$      $E_{a_{1}} = 28.31 \frac{kJ}{mol} = 28310 \frac{J}{mol}$

2) $C$ → $D$      $E_{a_{2}} = 20 \frac{kJ}{mol} = 20000\frac{J}{mol}$              

• Let $A_{1}, A_{2}$ and $k_{1}, k_{2}$ and $E_{a_{1}},E_{a_{2}}$ are the Arrhenius factor, rate constant  and activation energy for reaction 1 and reaction 2 respectively.

• Temperature both reaction $T= 400$ $K$

• Arrhenius Equation for reaction 1:

                                       $k_{1}=A_{1}\times e^\frac{-E_{a_{1}}}{RT}$

• Arrhenius Equation for reaction 2:

                                      $k_{2}=A_{2}\times e^\frac{-E_{a_{2}}}{RT}$

• The rate constants are equal. (Given)

                                     $k_{1}=k_{2}$

                           $A_{1} \times e^\frac{-E_{a_{1}}}{RT}=A_{2} \times e^\frac{-E_{a_{2}}}{RT}$

                                  $\frac{A_{1}}{A_{2}}= \frac{(e^{\frac{-E_{a_{2}}}{RT}})}{(e^{\frac{-E_{a_{1}}}{RT}})}$

                                       $= e^{\frac{-E_{a_{2}}}{RT}} \times e^{\frac{E_{a_{1}}}{RT}}$

                                  $\frac{A_{1}}{A_{2}}=e^{\frac{E_{a_{1}}-E_{a_{2}}}{RT}$

Substitute the values of $E_{a_{1}},E_{a_{2}},R$and $T$.

                                  $\frac{A_{1}}{A_{2}}=e^{\frac{E_{a_{1}}-E_{a_{2}}}{RT}$

                                       $= e ^{\frac{28310-20000}{8.314\times 400}}\\\\= e ^{\frac{8310}{33256}}\\\\= e^{ 2.4988}$

                                  $\frac{A_{1}}{A_{2}}=12.1679$

                                  $A_{1}=12.1679\times A_{2}$

                           $A_{1}:A_{2}=12.1679:1$

Hence, ratio of Arrhenius factor $A_{1}:A_{2}$ for reaction 1 and 2 $=12.1679:1$

To know more about the concept please go through the links

https://brainly.in/question/12223513

https://brainly.in/question/8159696

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