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Reika [66]
4 years ago
15

Explain how the electrical model for the control of cellvolume

Chemistry
1 answer:
AlexFokin [52]4 years ago
6 0

Answer:

Control of Cell Volume and Membrane Potential Basic

Basic problem

• The cell is full of components like Proteins, ions, fats, etc.

• Ordinarily, these would cause huge osmotic pressures, sucking water into the cell.

• The cell membrane has no structural strength, and the cell would burst.

Basic solution to the problem above

• Cells carefully regulate their intracellular ionic concentrations in order, to ensure that no osmotic pressures arise

• Consequently, the major ions Na+, K+, Cl- and Ca2+ in the cell have different concentrations in the extracellular and intracellular environments.

• And thus a voltage difference arises across the cell membrane.

Essentially two different kinds of cells: excitable and nonexcitable.

• All cells have a resting membrane potential, but only excitable cells modulate it actively.

Explanation:

In the case where there is a poisoning of the Na-K exchange pump.

Inhibition kicks off.

The pump requires binding by Na+ , K+ and ATP for its operation. Therefore , if the concentration of any of these substances is too low, the pump does not function.

When the temperature is reduced. During oxygen lack. Metabolic poisoning thus prevents the formation of ATP.

Potential buildup in the Na+ ion will be used by many different processes e.g cotransporter, neuron signalling etc.

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a) ΔHºrxn = 116.3 kJ, ΔGºrxn = 82.8 kJ,  ΔSºrxn =  0.113 kJ/K

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a)                                 C6H5−CH2CH3  ⇒  C6H5−CH=CH2  + H₂

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Sº J/mol                          255                          238                      130.6*

Note: This value was not given in our question, but is necessary and can be found in standard handbooks.

Using Hess law to calculate  ΔHºrxn we have

ΔHºrxn  = ΔHfº C6H5−CH=CH2 +  ΔHfº H₂ - ΔHºfC6H5−CH2CH3

ΔHºrxn =     103.8 kJ + 0 kJ  - (-12.5 kJ)

ΔHºrxn = 116.3 kJ

Similarly,

ΔGrxn = ΔGºf C6H5−CH=CH2 +  ΔGºfH₂ - ΔGºfC6H5CH2CH3

ΔGºrxn=   202.5 kJ + 0 kJ - 119.7 kJ  = 82.8 kJ

ΔSºrxn = 238 J/mol + 130.6 J/mol -255 J/K = 113.6 J/K = 0.113 kJ/K

b) The temperature at which the reaction is spontaneous or feasible occurs when ΔG becomes negative and using

ΔGrxn =  ΔHrxn -TΔS

we see that will happen when the term  TΔS  becomes greater than ΔHrxn since ΔS  is positive  , and so to sollve for T we will make ΔGrxn equal to zero and solve for T. Notice here we will make the assumption that  ΔºHrxn and ΔSºrxn remain constant at the higher temperature  and will equal the values previously calculated for them. Although this assumption is not entirely correct, it can be used.

0 = 116 kJ -T (0.113 kJ/K)

T = 1026.5 K  =  (1026.55 - 273 ) ºC = 753.55 ºC

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                       ΔGrxn =  ΔHrxn -TΔS

to calculate ΔGrxn   with the assumption that ΔHº and ΔSºremain constant.

ΔG =  116.3 kJ - (600+273 K) x 0.113 kJ/K =  116.3 kJ - 873 K x 0.113 kJ/K

ΔG =  116.3 kJ - 98.6 kJ =  17.65 kJ = 1.8 x 10⁴ J ( Note the kJ are converted to J to necessary for the next part of the problem )

Now for solving for K, the equation to use is

ΔG = -RTlnK and solve for K

- ΔG / RT = lnK  ∴ K = exp (- ΔG / RT)

K = exp ( - 1.8 x 10⁴ J /( 8.314 J/K  x 873 K)) = 8.2 x 10⁻²

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