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lions [1.4K]
3 years ago
11

gastric juice (ph 1.5) is produced by pumping HCl from blood plasma (ph 7.4) into the stomach. Calculate the amount of free ener

gy required to concentrate the H in 1 L of gastric juice at 37 degrees celsius
Chemistry
1 answer:
maxonik [38]3 years ago
7 0

Answer:

∆G = 35 KJ

Explanation:

The given data are as follows;

Gastric juice pH = 1.5

Blood plasma pH = 7.4

Temperature = (37 + 273) K = 310 K

However, pH = -log [H+]

Therefore, [H+] = 10^-pH

At pH of 1.5, [H+] = 10^-1.5 = 3.16 × 10^-2 M

At pH of 7.4, [H+] = 10^-7.4 = 3.98 × 10^-8 M

Using ∆G = RTln(C2/C1)

Where R = 8.314 J/mol/K;

T = 310 K;

C2 (concentration inside) = 3.16 × 10^-2 M;

C1 (concentration outside) = 3.98 × 10^-8 M

∆G = 8.314 × 310 × ln(3.16 × 10^-2 / 3.98 × 10^-8)

∆G = 35012.74

Therefore, ∆G = 35 KJ

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(LO 37,45) A reaction is 75% complete in 45.0 min. How long after its start will the reaction be 50% complete if it is (A) first
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Answer:

A) t = 22.5 min and B) t = 29.94 min

Explanation:

Initial concentration, [A]₀ = 100

Final concentration = 100 -75 = 25

Time = 45 min

A) First order reaction

ln[A] − ln[A]₀ = −kt

Solving for k;

ln[25] − ln[100] = - 45k

-1.386 = -45k

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Final concentration, [A] = 100 -50 = 50

Time = ?

ln[A] − ln[A]₀ = −kt

Solving for k;

ln[50] − ln[100] = - 0.0308 * t

-0.693 = -0.0308 * t

t = 22.5 min

B) Zero Order

[A] = [A]₀ − kt

Using the values from the initial reaction and solving for k, we have;

25 = 100 - k(45)

-75 = -45k

k = 1.67 M min-1

How long after its start will the reaction be 50% complete?

Initial concentration, [A]₀ = 100

Final concentration, [A] = 100 -50 = 50

Time = ?

[A] = [A]₀ − kt

50 = 100 - (1.67)t

-50 = - 1.67t

t = 29.94 min

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