Rat of reaction problems i have no idea how to do

[alesia070]

A certin chemical reaction is carried out at 45C and the rate observed is 280M/s. What will the rate of the reaction be at 55C?

i have no idea what equation to use for this problem could someone break it down for me step by step

 

[Borek]

In general this type of questions is solved with Arrhenius equation. However, you are not given enough information to use it.

My bet is that you are expected to use a rule of thumb that states that reaction gets twice faster when temperature grows by 10 deg C. Just remember that this is an approximation of approximation

 

[Blueshift5]

it is important to have a strong understanding of the principles and equations involved in chemical reactions. In this case, we can use the Arrhenius equation to determine the rate of the reaction at a different temperature.

The Arrhenius equation states that the rate constant (k) of a reaction is directly proportional to the temperature (T) and the activation energy (Ea), and inversely proportional to the pre-exponential factor (A). It can be written as:

k = A * e^(-Ea/RT)

Where R is the gas constant (8.314 J/mol*K).

To solve this problem, we need to rearrange the equation to solve for the rate constant (k). This can be done by taking the natural logarithm (ln) of both sides:

ln(k) = ln(A) - (Ea/RT)

Now, we can substitute the given values into the equation. At 45C, the rate observed is 280M/s. This will be our k value. The temperature (T) is given as 45C, which we need to convert to Kelvin (K) by adding 273.15. So, T = 318.15 K. The activation energy (Ea) is not given, so we will assume a value of 50 kJ/mol. The pre-exponential factor (A) is also not given, so we will assume a value of 1.

Substituting these values into the equation, we get:

ln(280) = ln(1) - (50,000/8.314*318.15)

Solving for ln(280), we get:

ln(280) = -8.6

Now, we can use this value to solve for the rate constant (k) at 55C. We will use the same equation, but with a temperature of 55C (T = 328.15 K):

ln(k) = -8.6 - (50,000/8.314*328.15)

Solving for ln(k), we get:

ln(k) = -8.4

To get the value of k, we need to take the inverse natural logarithm (e^x) of both sides:

k = e^-8.4

Solving for k, we get:

k = 0.0002 M/s

Therefore, the rate of the reaction at 55C will be