Answer: The activation energy Ea for this reaction is 22689.8 J/mol
Explanation:
According to Arrhenius equation with change in temperature, the formula is as follows.
![ln \frac{k_{2}}{k_{1}} = \frac{-E_{a}}{R}[\frac{1}{T_{2}} - \frac{1}{T_{1}}]](https://tex.z-dn.net/?f=ln%20%5Cfrac%7Bk_%7B2%7D%7D%7Bk_%7B1%7D%7D%20%3D%20%5Cfrac%7B-E_%7Ba%7D%7D%7BR%7D%5B%5Cfrac%7B1%7D%7BT_%7B2%7D%7D%20-%20%5Cfrac%7B1%7D%7BT_%7B1%7D%7D%5D)
= rate constant at temperature 
= 
= rate constant at temperature 
= 
= activation energy = ?
R= gas constant = 8.314 J/kmol
= temperature = 
= temperature = 
Putting in the values ::
![ln \frac{4.8\times 10^8}{2.3\times 10^8} = \frac{-E_{a}}{8.314}[\frac{1}{649} - \frac{1}{553}]](https://tex.z-dn.net/?f=ln%20%5Cfrac%7B4.8%5Ctimes%2010%5E8%7D%7B2.3%5Ctimes%2010%5E8%7D%20%3D%20%5Cfrac%7B-E_%7Ba%7D%7D%7B8.314%7D%5B%5Cfrac%7B1%7D%7B649%7D%20-%20%5Cfrac%7B1%7D%7B553%7D%5D)
The activation energy Ea for this reaction is 22689.8 J/mol
Answer:
When 0.250 moles of a gas is placed in a container at 25 °C, it exerts a pressure of 700 mm Hg.
Explanation:
Paramecia have a micro nuclei and a macro nuclei.
Answer:
4.42 × 10⁻³⁷ m
Explanation:
Step 1: Given and required data
- Mass of the body (m): 1 kg
- Velocity of the body (v): 1500 m/s
- Planck's constant (h): 6.63 × 10⁻³⁴ J.s
Step 2: Calculate the de Broglie wavelenght (λ) of the body
We will use de Broglie's equation.
λ = h / m × v
λ = (6.63 × 10⁻³⁴ J.s) / 1 kg × (1500 m/s) = 4.42 × 10⁻³⁷ m
