Question

The activation energy Ea for a particular reaction is 50.0 kJ/mol. How

much faster is the reaction at 322 K than at 310.0 K? (R = 8.314 J/mol
•K)

Answer 1

The reaction is approximately 2.081 times faster at 322 K than at 310 K.

Step-by-step explanation:

To determine how much faster a reaction is at one temperature versus another, we can use the Arrhenius equation, which is given by:

`\[ k = Ae^{-(E_a)/(RT)} \]`

where:
- k is the rate constant of the reaction
- A is the pre-exponential factor, which is a constant for a given reaction
- E_a is the activation energy
- R is the gas constant
- T is the temperature in Kelvin

To find out how much faster the reaction is at 322 K compared to 310 K, we need to find the rate constants at these temperatures (
`\(k_(322) \)` and
`\( k_(310) \)`), and then take their ratio. We won't need the value of A because we are interested in the ratio of rate constants, and A will cancel out.

Let's calculate
`\( k_(322) \)` and
`\( k_(310) \)` using the activation energy
`\( E_a = 50.0 \text{ kJ/mol} \)` and
`\( R = 8.314 \text{ J/mol•K} \)`. Remember to convert the activation energy to the same units as R, which are J/mol.

First, convert E_a to J/mol:

`\[ E_a = 50.0 \text{ kJ/mol} * 1000 \text{ J/kJ} = 50000 \text{ J/mol} \]`

Now, calculate k_{322}:

`\[ k_(322) = Ae^{-\frac{50000 \text{ J/mol}}{(8.314 \text{ J/mol•K})(322 \text{ K})}} \]`

Simplifying inside the exponent:

`\[ -(50000)/((8.314)(322)) \approx -18.691 \]`

Now,
`\( k_(322) = Ae^(-18.691) \)`

Next, calculate
`\( k_(310) \)`:

`\[ k_(310) = Ae^{-\frac{50000 \text{ J/mol}}{(8.314 \text{ J/mol•K})(310 \text{ K})}} \]`

Simplifying inside the exponent:

`\[ -(50000)/((8.314)(310)) \approx -19.424 \]`

So,
`\( k_(310) = Ae^(-19.424) \)`

To find out how much faster the reaction is at 322 K than at 310 K, we take the ratio of
`\( k_(322) \)` over
`\( k_(310) \)`:

`\[ (k_(322))/(k_(310)) = (Ae^(-18.691))/(Ae^(-19.424)) \]`

Because A is the same for both rate constants and we are only interested in the ratio, it cancels out:

`\[ (k_(322))/(k_(310)) = (e^(-18.691))/(e^(-19.424)) \]`

Using properties of exponents, we can combine the exponents:

`\[ (k_(322))/(k_(310)) = e^((-18.691) - (-19.424)) = e^(19.424 - 18.691) = e^(0.733) \]`

Using a calculator to find the value of
`\( e^(0.733) \)`:

`\[ (k_(322))/(k_(310)) \approx e^(0.733) \approx 2.081 \]`

This result means that the reaction is approximately 2.081 times faster at 322 K than at 310 K.

Answer 2

About two times faster.

Step-by-step explanation:

Hello!

In this case, since we need a k2/k1 ratio to figure out the increase in the reaction rate for a chemical process, we need to use the following version of the Arrhenius equation:

`ln((k_2)/(k_1) )=-(Ea)/(R)((1)/(T_2) -(1)/(T_1) )`

Thus, we plug in Ea, R and the temperatures to obtain:

`ln((k_2)/(k_1) )=-(50,000J/mol)/(8.314J/mol*K)((1)/(322K) -(1)/(310.0K) ) \\\\ln((k_2)/(k_1) )=0.723`

Now, we use exponential to obtain:

`(k_2)/(k_1) =exp(0.723)\\\\(k_2)/(k_1)=2.1`

Thus, we infer that the reaction is about two times faster.

Best regards!