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:
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If the two gases has a total pressure of 5.7 atm and one of the gases has a partial pressure of 4.1 the the other one has the pressure of 1.6
Answer:
E. potassium (K) and bromine (Br)
Explanation:
An ionic bond is formed between compounds with a large electronegativity difference between them. It is usually between a metal and non-metal.
- Potassium is a true metal found in group 1 on the periodic table.
- Bromine is a highly electronegative non-metal which is a halogen.
- Potassium will lose one of its electrons which will be gained by the Bromine.
- The electrostatic attraction between the two species will cause the ionic bond to form.
- The ability of one specie willing to lose electron and the other gaining, is the main bed rock of ionic bonding.
The ionization energy for a hydrogen atom in the n = 2 state is 328 kJ·mol⁻¹.
The <em>first ionization energy</em> of hydrogen is 1312.0 kJ·mol⁻¹.
Thus, H atoms in the <em>n</em> = 1 state have an energy of -1312.0 kJ·mol⁻¹ and an energy of 0 when <em>n</em> = ∞.
According to Bohr, Eₙ = k/<em>n</em>².
If <em>n</em> = 1, E₁= k/1² = k = -1312.0 kJ·mol⁻¹.
If <em>n</em> = 2, E₂ = k/2² = k/4 = (-1312.0 kJ·mol⁻¹)/4 = -328 kJ·mol⁻¹
∴ The ionization energy from <em>n</em> = 2 is 328 kJ·mol⁻¹
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