HomeworkLib
Question

Biochemistry: 90. The heat enthalpy change in a reaction where delta G is -9.92 kJ/mol and...

Biochemistry:
90. The heat enthalpy change in a reaction where delta G is -9.92 kJ/mol and delta S is -85.5 J/K at 25*C is:
a. 35.39 kJ/mol
b. -35.39 kJ/mol
c. 25.48 kJ/mol
d. -25.48 kJ/mol

science   chemistry   
0 0
Add a comment Improve this question Transcribed image text
Next > < Previous
Sort answers by oldest

Homework Answers

Request Professional Answer

Request Answer!

We need at least 10 more requests to produce the answer.

0 / 10 have requested this problem solution

The more requests, the faster the answer.

Request! (Login Required)

All students who have requested the answer will be notified once they are available.
Know the answer?
Add Answer to:
Biochemistry: 90. The heat enthalpy change in a reaction where delta G is -9.92 kJ/mol and...
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Log In

Sign Up

Similar Homework Help Questions

  • The change in enthalpy when 1 mol of ice is melted at 273K is 6008 J Heat Capacity of liquid water, Cp_{L} = 75.44 J/mol K Heat Capacity of solid water, Cp_{S} = 38J/mol K Enthalpy chage of melting at...

    The change in enthalpy when 1 mol of ice is melted at 273K is 6008 J Heat Capacity of liquid water, Cp_{L} = 75.44 J/mol K Heat Capacity of solid water, Cp_{S} = 38J/mol K Enthalpy chage of melting at 273K, \Delta H_{273}=6008 J Calculate the standard enthalpy of fusion for ice. Calculate the heat released when 100 g of water supercooled at 250K solidify Initial T=25°C=298K Thanks

  • For a given reaction, Delta H = -19.9 kJ/mol and Delta S = -55.5 J/K-mol. Calculate...

    For a given reaction, Delta H = -19.9 kJ/mol and Delta S = -55.5 J/K-mol. Calculate the temperature in K where Delta G = 0. Assume that Delta H and Delta S do not vary with temperature. Also, what is the equilibrium constant at that temperature?

  • What is the enthalpy change (KJ) during the process in which 200.0 g of water at...

    What is the enthalpy change (KJ) during the process in which 200.0 g of water at 60.0 Celsius is cooled to -25 Celsius? The specific heats of ice, and liquid water, and 2.03 J/g-K, and 4.18 J/g-K, respectively. For H 2 O, delta H fus = 6.01 kJ/mol

  • 1. The standard enthalpy change for the combustion of mole of propane is -2043.0 kJ. CH_(g)...

    1. The standard enthalpy change for the combustion of mole of propane is -2043.0 kJ. CH_(g) +5 0 (g) →3002(g) + 4H2O(g) Calculate Hº for propane based on the following standard molar enthalpies of formation. molecule CO (9) H2O(g) AH (kJ/mol-rxn) -393.5 -241.8 a. +1407.7 kJ/mol-rxn b. +104.7 kJ/mol-rxn c.-104.7 kJ/mol-rxn d -4190.7 kJ/mol-rxn e. -1407.7 kJ/mol-rxn 2. Which of the following statements is/are CORRECT? 1. If a reaction occurs at constant pressure, 4-AH. 2. The change in energy for...

  • For the reaction N_2(g) + 3 H_2(g) rightarrow 2 NH_3(g) Delta G degree = -23.6 kJ...

    For the reaction N_2(g) + 3 H_2(g) rightarrow 2 NH_3(g) Delta G degree = -23.6 kJ and Delta S degree = -198.7 J/K at 345 K and 1 atm. This reaction is (reactant, product) favored under standard conditions at 345 K. The standard enthalpy change for the reaction of 2.30 moles of N_2(g) at this temperature would be kJ. For the reaction 2 H_2O_2(l) rightarrow 2 H_2O(l) + O_2(g) Delta G degree = -236.9 kJ and Delta H degree =...

  • A reaction has an enthalpy change of - 71 kJ mol" and an entropy change of...

    A reaction has an enthalpy change of - 71 kJ mol" and an entropy change of -58 JK-mol-. At what temperature does this exothermic reaction cease to be spontaneous? 422°C 673°c 826°C O 951°C What is the molar reaction enthalpy for the reaction below: N2H4(0) +CHNO(0) 4CHẠO(g) +Ng(g) + 3H (g) given the following thermodynamic data? 2NH3(9) ► N2H4(l) + H2(9) A,Hm = 22.5 kJ mol-1 2NH3(9) N2(g) + 3H2(9) A Hm = 57.5 kJ mol-1 CH,0(9) + H2(g) +...

  • The heat (enthalpy change) of melting for gold is 12.5 kJ/mol, the entropy change for this...

    The heat (enthalpy change) of melting for gold is 12.5 kJ/mol, the entropy change for this process is 9.35 J/K mol. Calculate melting point for gold (hint use the formula for The Gibbs free energy change), beware units.  

  • For the reaction H2(g) + S(s) --> H2S(g) delta H = -20.2 kJ mol-1 and delta...

    For the reaction H2(g) + S(s) --> H2S(g) delta H = -20.2 kJ mol-1 and delta S =+43.1 J K-1mol-1. Which of the following statements is true? The reaction is spontaneous at all temperatures. delta G becomes less favorable as T is raised. The reaction is only spontaneous at high temps. The reaction is only spontaneous at low temps. The reaction is at equilibrium at 25 C under standardconditions. Please explain why too. Thank you, feedback will beawarded as soon...

  • 8.) From the values of delta H and delta S, calculate delta G then predict whether...

    8.) From the values of delta H and delta S, calculate delta G then predict whether the following reactions would be spontaneous or not at 25 C. a) Reaction A: delta H= 10.5 kJ/mol, and delta S = 30 J/K mol b) Reaction B: delta H=1.8 kJ/mol, and delta S = -113 J/K mol 9.) Calculate the delta G and K, for the following equilibrium reaction at 25 C: 2H2O(Ⓡ) <-> 2H2(g) + O2(8) delta Gran H2O(x) = -228.6 kJ/mol

  • Data: SC2(g) + 1/2(g) = S03(g) delta H = -99 kJ/mol Use the data above to...

    Data: SC2(g) + 1/2(g) = S03(g) delta H = -99 kJ/mol Use the data above to calculate the standard enthalpy change for the reaction below Reaction: 2 So_3(g) = 02(g) + 2 S02(g) delta H = ? kJ/mol a-198 kJ/mol B. 99 kJ/mol C.-99 kJ/mol D-198 kJ/mol Q E. 49.5 kJ/mol

ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT