How might the yield of 1-bromobutane be affected if water was not added, and what product(s) would be favored?
1 answer:
- Due to the inability of the reaction to take place, the yield of 1-Bromobutane would drop.
- Since 1-Butanol won't react with the additional sodium bromide, bromination won't happen.
- If water had been supplied, the equilibrium would have shifted extremely far to the left, preventing the reactants from interacting with the acid and favoring the yield of 1-bromobutane instead.
- When a substance undergoes bromination, bromine is added to the compound as a result of the chemical reaction.
- After bromination, the result will have different properties from the initial reactant.
- For example, an alkene is brominated by electrophilic addition of
.
- Benzene ring bromination by electrophilic aromatic substitution.
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Answer: The partial pressure of is 1.86 atm
Explanation:
Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as
The given balanced equilibrium reaction is,
Pressure at eqm. 0.973 atm 0.548atm x atm
The expression for equilibrium constant for this reaction will be,
Now put all the given values in this expression, we get :
By solving the term 'x', we get :
x = 1.86 atm
Thus, the partial pressure of is 1.86 atm
Answer:
The amount of energy required to transport hydrogen ions from a cell into the stomach is 37.26KJ/mol.
Explanation:
The free change for the process can be written in terms of its equilibrium constant as:
ΔG° =
where:
R= universal gas constant
T= temperature
= equilibrum constant for the process
Similarly, free energy change and cell potentia; are related to each other as follows;
ΔG= -nFE°
from above;
F = faraday's constant
n = number of electrons exchanged in the process; and
E = standard cell potential
∴ The amount of energy required for transport of hydrogen ions from a cell into stomach lumen can be calculated as:
ΔG° =
where;
[texK_eq[/tex]=
For transport of ions to an internal pH of 7.4, the transport taking place can be given as:
⇒
Equilibrum constant for the transport is given as:
= 10⁻⁷⁴
=3.98 * 10⁻⁸M
= 10⁻²¹
=7.94 * 10⁻³M
Hence;
=
= 5.012 × 10⁻⁶
Furthermore, free energy change for this reaction is related to the equilibrium concentration given as:
ΔG° =
If temperature T= 37° C ; in kelvin
=37° C + 273.15K
=310.15K; and
R-= 8.314 j/mol/k
substituting the values into the equation we have;
ΔG₁ =
= 31467.93Jmol⁻¹
≅ 31.47KJmol⁻¹
If the potential difference across the cell membrane= 60.0mV.
Energy required to cross the cell membrane will be:
ΔG₂ =
ΔG₂ =
= 5.79KJ
Therefore, for one mole of electron transfer across the membrane; the energy required is 5.79KJmol⁻¹
Now, we can calculate the total amount of energyy required to transport H⁺ ions across the membrane:
Δ
= (31.47+5.79) KJmol⁻¹
= 37.26KJmol⁻¹
We can therefore conclude that;
The amount of energy required to transport ions from cell to stomach lumen is 37.26KJmol⁻¹