What is the energy requirement in kJ/mol to transport a proton across the mitochondrial inner membrane in plant cells at night when the outside temperature is 15ºC, the pH differential across this membrane is 1.4 pH units, and the membrane potential is 160 mV? Choose the ONE correct answer.

A. +21.8 kJ/mol
B. +75.4 kJ/mol
C. +23.1 kJ/mol
D. +46.2 kJ/mol
E. -21.8 kJ/mol
F. -23.1 kJ/mol
G. +19.0 kJ/mol
H. +44.6 kJ/mol

Question

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Answer :

ajeigbeibraheem ajeigbeibraheem · Answer 1 · 2024-04-03T08:20:52-04:00

Answer:

C. +23.1 kJ/mol

Explanation:

the formula to use to calculate the energy requirement in kJ/mol to transport a proton across the mitochondrial inner membrane in plant cells is:

ΔGt = RTIn [tex]\frac{C_{2} }{C_1}[/tex] + ZFΔV

let's list the values of the data we are being given in the question to make it easier when solving it.

Z= 1

F= 96500C (faraday's constant)

ΔV= 160mV = 0.160V

R= 8.314( constant)

T= 15ºC ( converting our degree Celsius into kelvin, we will have 273.15k+ 15 = 288.15K)

∴ T= 288.15K

Putting it all together in the formula, we have:

ΔGt = 8.314 × 288.15 × 2.303 log [tex]\frac{C_{2} }{C_1}[/tex] + 1 × 96500 × 0.160

ΔGt = 5517.25 [tex][ -(log(H^{+}_{out} ) + log(H^{+}_{in} )_{in}[/tex] +15440

ΔGt = 5517.25 [tex](pH_{out} - pH_{in}[/tex] +15440

Given that the pH differential gradient across the membrane is 1.4pH units. It implies that;

ΔGt = 5517.25 × 1.4 + 15440

= 7724.15 +15440

= 23164.15 Joules/moles

= +23.1 KJ/mole