What is the equilibrium membrane potential due to na ions if the extracellular concentration of na ions is 142 mm and the intracellular concentration of na ions is 27 mm at 20°c?

The equilibrium membrane potential is 41.9 mV.

To calculate the membrane potential, we use the Nernst Equation:

V_Na = (RT)/(zF) ln{[Na]_o/[Na]_ i}

where

•V_Na = the equilibrium membrane potential due to the sodium ions

•R = the universal gas constant [8.314 J·K^(-1)mol^(-1)]

•T = the Kelvin temperature

•z = the charge on the ion (+1)

•F = the Faraday constant [96 485  C·mol^(-1) = 96 485 J·V^(-1)mol^(-1)]

•[Na]_o = the concentration of Na^(+) outside the cell

•[Na]_i = the concentration of Na^(+) inside the cell

∴ V_Na =

[8.314 J·K^(-1)mol^(-1) × 293.15 K]/[1 × 96 485 J·V^(-1)mol^(-1)] ln(142 mM/27 mM) = 0.025 26 V × ln5.26 = 1.66× 25.26 mV = 41.9 mV

the potential energy of the reactants is 200j.

from the energy diagram, the energy of the product formed is 350j; this means that, this reaction is an endothermic reaction, because it absorbs energy from its environment.