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Nataly_w [17]
1 year ago
14

Imagine that a researcher has developed a way to isolate thylakoids from chloroplasts and to make the ph of the thylakoid space

acidic. If these isolated thylakoids are transferred to a ph8 solution in the dark, what might the researcher expect to observe?.
Chemistry
1 answer:
PilotLPTM [1.2K]1 year ago
8 0

In the given condition where the thylakoids are isolated from the chloroplasts to make the pH of the thylakoid space acidic, the isolated thylakoid will form the ATP.

ATP is also referred to as Adenosine Triphosphate and is correctly called as the energy currency of the cell as it is responsible molecule for carrying the energy units in the cell. When the isolated thylakoids are made to transfer in the pH = 8 basic solution, then the acidic lumen of the chloroplast will be having a higher concentration of H⁺ ions. When they are placed in the basic solution then the H⁺ ions will move out because of the occurrence of concentration gradient which is higher in the lumen than the surrounding. The chloroplasts will now bear synthesis of ATP. Therefore, the isolated thylakoid is able to make ATP when the lumen is acidic, and the surroundings are made basic in nature.

Thus, the thylakoid which is isolated will bear the synthesis of ATP.

Learn more about ATP at:

brainly.com/question/252380

#SPJ9

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vladimir1956 [14]

Answer:

Kp^{1000K}=0.141\\Kp^{298.15K}=2.01x10^{-18}

\Delta _rG=1.01x10^5J/mol

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

C_2H_6(g)\rightleftharpoons H_2(g)+C_2H_4(g)

Thus, Kp for this reaction is computed based on the given molar fractions and the total pressure at equilibrium, as shown below:

p_{C_2H_6}^{EQ}=2bar*0.592=1.184bar\\p_{C_2H_4}^{EQ}=2bar*0.204=0.408bar\\p_{H_2}^{EQ}=2bar*0.204=0.408bar

Kp=\frac{p_{C_2H_4}^{EQ}p_{H_2}^{EQ}}{p_{C_2H_6}^{EQ}}=\frac{(0.408)(0.408)}{1.184}=0.141

Now, by using the Van't Hoff equation one computes the equilibrium constant at 298.15K assuming the enthalpy of reaction remains constant:

Ln(Kp^{298.15K})=Ln(Kp^{1000K})-\frac{\Delta _rH}{R}*(\frac{1}{298.15K}-\frac{1}{1000K} )\\\\Ln(Kp^{298.15K})=Ln(0.141)-\frac{137000J/mol}{8.314J/mol*K} *(\frac{1}{298.15K}-\frac{1}{1000K} )\\\\Ln(Kp^{298.15K})=-40.749\\\\Kp^{298.15K}=exp(-40.749)=2.01x10^{-18}

Finally, the Gibbs free energy for the reaction at 298.15K is:

\Delta _rG=-RTln(Kp^{298.15K})=8.314J/mol*K*298.15K*ln(2.01x10^{-18})\\\Delta _rG=1.01x10^5J/mol

Best regards.

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