# Given that the initial rate constant is 0.0140s-1 at an initial temperature of 29 C , what would...

## Question:

Given that the initial rate constant is 0.0140s-1 at an initial temperature of 29 C , what would the rate constant be at a temperature of 170. C for the same reaction described in Part A?

Part A was the activation energy of a certain reaction is 46.7 kJ/mol . At 29 C , the rate constant is 0.0140s?1. At what temperature in degrees Celsius would this reaction go twice as fast? T2= 41C

## Rate Constants

Rate constants have a misleading name, since they are not constant. They depend on temperature, reaction thermodynamics (like activation energy), and the molecular kinetics of the mixture. The relevant equation is the Arrhenius equation:

{eq}k = Ae^{\dfrac{-E_a}{RT}} {/eq}

Become a Study.com member to unlock this answer!

The activation energy is given, and we're only missing the final rate constant. We can find it like this:

{eq}T_1 = 302K\\ T_2 =...

Rate Constant and Rate Laws

from

Chapter 12 / Lesson 2
17K

Learn the difference between rate constant and rate law. Explore how to use the rate law equation to find the reaction order for one and two reactants.