For a particular reaction at 129.9 °C, ΔG = 319.61 kJ/mol, and ΔS = 777.00 J/(mol·K)....

For a particular reaction at 129.9 °C, ΔG = 319.61 kJ/mol, and ΔS = 777.00 J/(mol·K). delta g is -47.6 celcius

Solutions

Expert Solution

T = 129.9 + 273 = 402.9 K

ΔG = 319.61 kJ/mol

ΔG = - RT ln K

319.61 = -8.314 x 10^-3 x 402.9 x ln K

lnK = -95.41

K = 3.66 x 10^-42

ΔS = 777.00 J/(mol·K)

= 0.777 kJ / mol K

ΔG = ΔH - T ΔS

319.61 = ΔH - 402.9 x 0.777

ΔH = 632.66 kJ / mol K

now :

T2 = 402.9 K

K2 = 3.66 x 10^-42

T1 = -47.6+273 = 225.4 K

K1 = ?

ln (K2 / K1) = ΔH / R ) [1/T1 - 1/T2]

ln (3.66 x 10^-42 / K1 ) = 632.66 / (8.314 x 10^-3) [1/225.4 - 1/402.9]

= 148.73

K1 = 9.36 x 10^-107

ΔG = - RT ln K

ΔG = -8.314 x 10^-3 x 225.4 x ln (9.36 x 10^-107)

ΔG = 457.51 kJ / mol

queen_honey_blossom answered 1 year ago

Next > < Previous

Related Solutions

For a particular reaction at 234.1 °C, ΔG = -832.94 kJ/mol, and ΔS = 923.32 J/(mol·K)....

For a particular reaction at 234.1 °C, ΔG = -832.94 kJ/mol, and ΔS = 923.32 J/(mol·K). Calculate ΔG for this reaction at -105.7 °C.

For a particular reaction at 215.7 °C, ΔG = 404.91 kJ/mol, and ΔS = 748.20 J/(mol·K)....

For a particular reaction at 215.7 °C, ΔG = 404.91 kJ/mol, and ΔS = 748.20 J/(mol·K). Calculate delta G for -74.3 celcius. I've asked this question previously and the individual who answered got roughly 378 kj/mol which is incorrect.

For a particular reaction at 129.9 °C, ΔG = -603.84 kJ/mol, and ΔS = 313.84 J/(mol·K).Calculate...

For a particular reaction at 129.9 °C, ΔG = -603.84 kJ/mol, and ΔS = 313.84 J/(mol·K).Calculate ΔG for this reaction at -3.4 °C.

c. For this reaction in heart muscle, ΔG°’ = +4.7 kJ/mol but ΔG = -0.6 kJ/mol....

c. For this reaction in heart muscle, ΔG°’ = +4.7 kJ/mol but ΔG = -0.6 kJ/mol. i. Explain, in words, how ΔG can be negative when ΔG°’ is positive. ii. What would be the ratio of 3PG to 2PG if the reaction were at equilibrium at 25°C? iii. What is the actual ratio of 3PG to 2PG in heart muscle (T = 37°C)?

1. Reaction 1 has a ΔG° of –12.3 kJ/mol, and Reaction 2 has a ΔG° of...

1. Reaction 1 has a ΔG° of –12.3 kJ/mol, and Reaction 2 has a ΔG° of 23.4 kJ/mol. Which statement is TRUE of these two reactions? Select one: A. Reaction 1 occurs faster. B. Reaction 2 occurs faster. C. Both reactions occur at the same rate. D. Reaction 2 will not occur. E. It is impossible to know which reaction occurs faster with this information. 2. Fructose-1-phosphate can be hydrolyzed into fructose + inorganic phosphate (Pi) with a ΔG° of...

Given that ΔG∘ = −13.6 kJ/mol, calculate ΔG at 25∘C for the following sets of conditions....

Given that ΔG∘ = −13.6 kJ/mol, calculate ΔG at 25∘C for the following sets of conditions. 1) 30 atm NH3, 30 atm CO2, 4.0 M NH2CONH2 Express the free energy in kilojoules per mole to two significant figures. 2) 8.0×10−2 atm NH3, 8.0×10−2 atm CO2, 1.0 M NH2CONH2 Express the free energy in kilojoules per mole to two significant figures. Is the reaction spontaneous for the conditions in Part A and/or Part B? A) Is the reaction spontaneous for the...

What is the intracellular glucose concentration if the ΔG for the following reaction is -20.1 kJ/mol...

What is the intracellular glucose concentration if the ΔG for the following reaction is -20.1 kJ/mol at 37°C and concentrations for glucose-6-phosphate and phosphate are both 1 mM? glucose-6-phosphate --------> glucose + Pi ΔG° = -13.8 kJ/mol A) 1.9 M B) 87 M C) 1.9 mM D) 27 mM E) 87 mM The Answer is E. I just need help understanding why it is E. please show all your work. I really need to understand the entire process. Please don't...

Use data in Appendix C in the textbook to calculate ΔH∘ in (kJ/mol) , ΔS∘ in...

Use data in Appendix C in the textbook to calculate ΔH∘ in (kJ/mol) , ΔS∘ in (j/mol-K) , and ΔG∘ in (kJ/mol) at 25 ∘C for each of the following reactions: 2P(g)+10HF(g)→2PF5(g)+5H2(g) Appendix C: P(g): dH:316.4; dG: 280.0; S:163.2 HF(g): dH: -268.61; dG: -270.70; S:173.51 PF5(g): dH:-1594.4; dG: -1520.7; S: 300.8 H2(g): dH:217.94; dG: 203.26; S: 114.60

Consider a reaction with the following thermodynamic properties. ΔH° = 77.7 kJ ΔS° = –35.7 J/(K...

Consider a reaction with the following thermodynamic properties. ΔH° = 77.7 kJ ΔS° = –35.7 J/(K • mol) ΔG°= 88.4 kJ This reaction: will proceed very slowly. will be spontaneous at low temperatures. has bonds in the products that are weaker than the reactants. may have fewer and more complicated molecules in the product.

For a hypothetical reaction, ΔrHθ = 89.1 kJ/mol and ΔrSθ = 364 J/mol/K. What is the...

For a hypothetical reaction, ΔrHθ = 89.1 kJ/mol and ΔrSθ = 364 J/mol/K. What is the total standard entropy of the reaction, ΔrStotθ, at 343 K?

Subjects