Chemistry, 11.04.2020 01:50 ariellllllllllllllll

The equation for ATP hydrolysis is
ATP yields ADP + Pi delta G*= -30.5 kJ/mol

( H20)

(a) Calculate ?G for ATP hydrolysis to rank the following conditions from most favorable to least favorable. Assume a temperature of 37.0C. R = 8.315 J/(mol. K).
ATP Hydrolysis most favorable
ATP hydroysis least favorable
The choices are:

a. muscle: [ATP]= 8.1mM; [ADP]= 0.9mM [Pi]= 8.1mM
b. brain: [ATP]= 2.6mM; [ADP]= 0.7mM [Pi]= 2.7mM
c. liver: [ATP]= 3.4mM; [ADP]= 1.3mM [Pi]= 4.8mM

(b) Calculate ?G for ATP hydrolysis in muscle at 18 degree C. Use the muscle concentrations from part a.
Delta g = kJ/mol

Answers: 2

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Answer from: kiannadgarnica

The most favorable is the brain tissue, followed by the muscle tissue while the least favorable is the liver tissue.

The equation of the reaction is; ATP > ADP + Pi ΔG°= -30.5 kJ/mol

For muscle tissue;

Q = [ADP] [Pi]/[ATP]

Q = [0.9 × 10^-3] [8.1 × 10^-3]/[[8.1 × 10^-3]

Q = 9.0 × 10^-4

ΔG =ΔG° + RTlnQ

Where ΔG°= -30.5 kJ/mol or -30.5 × 10^3 J/mol from the equation

ΔG = -30.5 kJ/mol + (8.315 × 310 × ln(9.0 × 10^-4)) kJ/mol

ΔG =-48.6 kJ/mol

For the brain tissue;

Q = [0.7 × 10^-3] [2.7 × 10^-3]/[2.6 × 10^-3]

Q = 7.3 × 10^-4

ΔG = -30.5 kJ/mol + (8.315 × 310 × ln(7.3 × 10^-4)) kJ/mol

ΔG = -49.1 kJ/mol

For the liver tissue;

Q = [1.3 × 10^-3] [4.8 × 10^-3]/[3.4 × 10^-3]

Q = 1.8 × 10^-3

ΔG = -30.5 kJ/mol + (8.315 × 310 × ln( 1.8 × 10^-3)) kJ/mol

ΔG = -46.3 kJ/mol

The most favorable is the brain tissue, followed by the muscle tissue while the least favorable is the liver tissue.

For the muscle tissue at 18°;

ΔG = -30.5 kJ/mol + (8.315 × 291 × ln(9.0 × 10^-4)) kJ/mol

ΔG = -47.5 kJ/mol

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Answer from: kledi72

Here's what I get

Explanation:

ATP ⟶ ADP + Pi; ΔG° = -30.5 kJ·mol⁻¹

(a) Rank the favourability of ATP hydrolysis

To get ΔG under physiological conditions, we must use the equation

ΔG = ΔG° + RTlnQ where

$Q = \dfrac{\text{[ADP][P$_{\text{i}}$]}}{\text{[ADP]}}$

(i) In muscle tissue

$Q = \dfrac{0.9 \times 10^{-3} \times 8.1 \times 10^{-3}}{8.1 \times 10^{-3}} = 0.9 \times 10^{-3}\\\\RT \ln Q = 8.315 \times 310.15 \times \ln (0.9 \times 10^{-3}) = 2579 \times (-7.01) = \text{-18 100 J/mol} = \text{-18.1 kJ/mol}\\\Delta G = \Delta G^{\circ} + RT \ln Q = -30.5 -18.1 = \textbf{-48.6 kJ/mol}$

(ii) In brain tissue

$Q = \dfrac{0.7 \times 10^{-3} \times 2.7 \times 10^{-3}}{2.6 \times 10^{-3}} = 0.73 \times 10^{-3}\\\\RT \ln Q = 8.315 \times 310.15 \times \ln (0.73 \times 10^{-3}) = 2579 \times (-7.23) = \text{-18 100 J/mol} = \text{-18.6 kJ/mol}\\\Delta G = \Delta G^{\circ} + RT \ln Q = -30.5 -18.6 = \textbf{-49.1 kJ$\cdot$mol$^{\mathbf{-1}}$}$

(iii) In liver tissue

$Q = \dfrac{1.3 \times 10^{-3} \times 4.8 \times 10^{-3}}{3.4 \times 10^{-3}} = 1.84 \times 10^{-3}\\\\RT \ln Q = 8.315 \times 310.15 \times \ln (0.73 \times 10^{-3}) = 2579 \times (-6.30) = \text{-16 200 J/mol} = \text{-16.2 kJ/mol}\\\Delta G = \Delta G^{\circ} + RT \ln Q = -30.5 -16.2 = \textbf{-46.7 kJ$\cdot$mol$^{\mathbf{-1}}$}$

$\begin{array}{cc}\textbf{Relative}\\\textbf{Favourability} & \textbf{Tissue}\\1& \text{brain (-49.1 kJ/mol)}\\2& \text{muscle (-48.6 kJ/mol)}\\3& \text{liver (-46.7 kJ/mol)}\\\end{array}$

(b) ΔG in muscle at 18 °C

T= (18.0 + 273.15) = 291.15 K

$Q = \dfrac{0.9 \times 10^{-3} \times 8.1 \times 10^{-3}}{8.1 \times 10^{-3}} = 0.9 \times 10^{-3}\\\\RT \ln Q = 8.315 \times 291.15 \times \ln (0.9 \times 10^{-3}) = 2421\times (-7.01) = \text{-17 000 J/mol} = \text{-17.0 kJ/mol}\\\Delta G = \Delta G^{\circ} + RT \ln Q = -30.5 -17.0 = \textbf{-47.5 kJ$\cdot$mol}^{\mathbf{-1}}}$

Answer from: Quest

specific gravity is the density of a substance divided by the density of water. since (at standard temperature and pressure) water has a density of 1 gram/cm3, and since all of the units cancel, specific gravity is usually very close to the same value as density

so "c" is the answer

plz give brainliest

Answer from: Quest

Afraidsugar lol it’s the third one

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The equation for ATP hydrolysis is
ATP yields ADP + Pi delta G*= -30.5 kJ/mol

( H2...