Question

The following data show the rate constant of a reaction measured at several different temperatures. Temperature (K) Rate Constant (1/s) 310 0.194 320 0.554 330 1.48 340 3.74 350 8.97 Part APart complete Use an Arrhenius plot to determine the activation barrier for the reaction. Express your answer using three significant figures. Ea

Answer 1

To determine the activation barrier for the reaction, we can use an Arrhenius plot. The Arrhenius equation relates the rate constant of a reaction to the temperature and the activation energy. By taking the natural logarithm of the rate constant and plotting it against the inverse of the temperature (in Kelvin), we can obtain a linear graph.

Step-by-step explanation:

To determine the activation barrier for the reaction, we can use an Arrhenius plot. The Arrhenius equation relates the rate constant of a reaction to the temperature and the activation energy. By taking the natural logarithm of the rate constant and plotting it against the inverse of the temperature (in Kelvin), we can obtain a linear graph. The slope of this graph is equal to the negative of the activation energy divided by the gas constant. Therefore, by calculating the slope of the Arrhenius plot, we can determine the value of the activation energy for the reaction.

Answer 2

Complete Question

The complete question is shown on the first uploaded image

Answer:

Part A

activation barrier for the reaction
`E_a = 84 .0 \ KJ/mol`

Part B

The frequency plot is
`A = 2.4*10^(13) s^(-1)`

Step-by-step explanation:

From the question we are told that

at
`T_1 = 300 \ K`
`k_1 = 5.70 *10^(-2)`

and at
`T_2 = 310 \ K`
`k_2 = 0.169`

The Arrhenius plot is mathematically represented as

`ln [(k_2)/(k_1) ] = (E_a)/(R) [(1)/(T_1) - (1)/(T_2) ]`

Where
`E_a` is the activation barrier for the reaction

R is the gas constant with a value of
`R = 8.314*10^(-3) KJ/mol \cdot K`

Substituting values

`ln [\frac{0.169}{6*10^-2{}} ] = (E_a)/(8.314*10^(-3)) [(1)/(300) - (1)/(310) ]`

=>
`E_a = 84 .0 \ KJ/mol`

The Arrhenius plot can also be mathematically represented as

`k = A * e^{-(E_a)/(RT) }`

Here we can use any value of k from the data table with there corresponding temperature let take
`k_2 \ and \ T_2`

So substituting values

`0.169 = A e ^{- (84.0)/(8.314*10^(-3) * 310) }`

=>
`A = 2.4*10^(13) s^(-1)`