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1 year ago

What is the equilibrium constant for the following reaction at 25 °c?

Physics

1 answer:

sdas [7]1 year ago

4 0

The equilibrium constant for the provided reaction at the temperature of 25 degree Celsius is 1/20.

<h3>What is equilibrium constant?</h3>

The equilibrium constant in a chemical reaction is the reaction quotient when the chemical reaction is at equilibrium constant.

The equilibrium constant for the for reaction at 25 °c given in the problem as,

$2A\rightleftharpoons B+C, K_1=1.0 \\2B\rightleftharpoons C+D, K_2=16 \\2C+D\rightleftharpoons 2P, K_3=25 \\$

Now, the reaction for which the equilibrium constant at 25 °c has to be find out is,

$P\rightleftharpoons A+\dfrac{1}{2}B$

$K_4=\sqrt{\dfrac{1}{K_1}+\dfrac{1}{K_2}+\dfrac{1}{K_3}}\\K_4=\sqrt{\dfrac{1}{25}+\dfrac{1}{16}+\dfrac{1}{1}}\\K_4=\dfrac{1}{20}$

Thus, the equilibrium constant for the provided reaction at the temperature of 25 degree Celsius is 1/20.

Learn more about the equilibrium constant here;

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3 years ago

Read 2 more answers

The membrane that surrounds a certain type of living cell has a surface area of 5.1 x 10-9 m2 and a thickness of 1.4 x 10-8 m.

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a) The charge on the outer surface is $1.2\cdot 10^{-12} C$

b) The number of ions is $7.5\cdot 10^6$

Explanation:

The membrane behaves as a parallel plate capacitor, whose capacity is given by the equation

$C=\frac{k\epsilon_0 A}{d}$

where

k = 4.3 is the dielectric constant

$\epsilon_0 =8.85\cdot 10^{-12} F/m$ is the vacuum permittivity

$A=5.1\cdot 10^{-9} m^2$ is the surface area

$d=1.4\cdot 10^{-8} m$ is the distance between the two plates

Substituting,

$C=\frac{(4.3)(8.85\cdot 10^{-12})(5.1\cdot 10^{-9})}{1.4\cdot 10^{-8}}=1.4\cdot 10^{-11} F$

The capacity of the membrane is related to the potential difference between the two surfaces by

$C=\frac{Q}{\Delta V}$

where here we have

Q = excess charge on one surface

$\Delta V = 85.5 mV = 0.0855 V$ is the potential difference between the two surfaces

Solving for Q, we find

$Q=C\Delta V=(1.4\cdot 10^{-11})(0.0855)=1.2\cdot 10^{-12} C$

We said that the net charge on the outer surface is

$Q=1.2\cdot 10^{-12} C$

The charge of one K+ ions is equal to the electron charge

$+e=1.6\cdot 10^{-19} C$

Therefore, the number of ions on the outer surface can be found by dividing the total charge by the charge of a single ion:

$N=\frac{Q}{e}=\frac{1.2\cdot 10^{-12}}{1.6\cdot 10^{-19}}=7.5\cdot 10^6$

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