asked 68.6k views
2 votes
A reaction has rate constant of 0.000122 at 27 and 0.228 at 77

1 Answer

6 votes

Answer:

The information you've provided seems to be related to a chemical reaction and its rate constants at two different temperatures: 27°C and 77°C. The rate constant (often denoted as k) for a chemical reaction typically follows the Arrhenius equation, which relates the rate constant to temperature:

\[k = A \cdot e^{\frac{-E_a}{RT}}\]

Where:

- \(k\) is the rate constant.

- \(A\) is the pre-exponential factor.

- \(E_a\) is the activation energy.

- \(R\) is the gas constant (8.314 J/(mol·K)).

- \(T\) is the absolute temperature in Kelvin.

In your case, you have two sets of data:

1. At 27°C (300 Kelvin), \(k = 0.000122\).

2. At 77°C (350 Kelvin), \(k = 0.228\).

You can use these two data points to calculate the activation energy (E_a) and the pre-exponential factor (A) for this reaction. To do this, you'll need to rearrange the Arrhenius equation and solve for \(E_a\) and \(A\):

1. Use the first data point (27°C):

\[0.000122 = A \cdot e^{\frac{-E_a}{(8.314 \cdot 300)}}\]

2. Use the second data point (77°C):

\[0.228 = A \cdot e^{\frac{-E_a}{(8.314 \cdot 350)}}\]

By solving these two equations simultaneously, you can determine the values of \(E_a\) and \(A\) for the reaction. This will give you more insight into the kinetics of the reaction at these temperatures.

answered
User Artuska
by
7.3k points

No related questions found

Welcome to Qamnty — a place to ask, share, and grow together. Join our community and get real answers from real people.