Question

Carbon disulfide (CS2) is a toxic, highly flammable substance. The following thermodynamic data are available for...

Carbon disulfide (CS2) is a toxic, highly flammable substance. The following thermodynamic data are available for CS2(l) and CS2(g) at 298 K:

ΔHf (kJ/mol) ΔGf (kJ/mol)
CS2(l) 89.7 65.3
CS2(g) 117.4 67.2

1. Liquid CS2 burns in O2 with a blue flame, forming CO2(g) and SO2(g). Write a balanced equation for this reaction.

2. Using the data in the preceding table and in Appendix C in the textbook, calculate ΔH∘ and ΔG∘ for the reaction in part D.

3. Use the data in the preceding table to calculate ΔS∘ at 298 K for the vaporization of CS2(l).

4. Using data in the preceding table and your answer to part 3, estimate the boiling point of CS2(l).

Homework Answers

Answer #1

1. Carbon disulfide is highly toxic and flammable substance. When it burns with oxygen it produces carbon dioxide and Sulfur dioxide.

The chemical reaction can be written as

The balanced reaction is written as:

Now we have C, S, O atoms in overall reaction.

After balancing , the number of oxygen atoms at reactant side 6 as well as at product side 6

The number of C atom at reactant side is 1 and at product side is also 1

The number of Sulfur atom at reactant side is 2 and also at product side is 2.

This answer of question number 1.

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
he thermodynamic properties for a reaction are related by the equation that defines the standard free...
he thermodynamic properties for a reaction are related by the equation that defines the standard free energy, ΔG∘, in kJ/mol: ΔG∘=ΔH∘−TΔS∘ where ΔH∘ is the standard enthalpy change in kJ/mol and ΔS∘ is the standard entropy change in J/(mol⋅K). A good approximation of the free energy change at other temperatures, ΔGT, can also be obtained by utilizing this equation and assuming enthalpy (ΔH∘) and entropy (ΔS∘) change little with temperature. Part A For the reaction of oxygen and nitrogen to...
1- Calculate ΔG o for the following reaction at 25°C. You will have to look up...
1- Calculate ΔG o for the following reaction at 25°C. You will have to look up the thermodynamic data. 2 C2H6(g) + 7 O2(g) → 4 CO2(g) + 6 H2O(l) 2- A reaction will be spontaneous only at low temperatures if both ΔH and ΔS are negative. For a reaction in which ΔH = −320.1 kJ/mol and ΔS = −99.00 J/K ·mol,determine the temperature (in °C)below which the reaction is spontaneous.
1)Use standard thermodynamic data (in the Chemistry References) to calculate G at 298.15 K for the...
1)Use standard thermodynamic data (in the Chemistry References) to calculate G at 298.15 K for the following reaction, assuming that all gases have a pressure of 19.31 mm Hg. 2N2(g) + O2(g)2N2O(g) G = ? kJ/mol 2)Using standard thermodynamic data (linked), calculate the equilibrium constant at 298.15 K for the following reaction. C2H4(g) + H2O(g)CH3CH2OH(g) K = ? 3) Calculate the temperature (in kelvins) at which the sign of G° changes from positive to negative for the reaction below. This...
For the gaseous reaction of xenon (Xe) and fluorine (F) to form xenon hexafluoride (XeF6), ΔH°...
For the gaseous reaction of xenon (Xe) and fluorine (F) to form xenon hexafluoride (XeF6), ΔH° = -402 kJ/mol and ΔG° = -280 kJ/mol at 298 K. Xe (g) + 3F2 (g) → XeF6 (g) a. Calculate the ΔS° for the reaction. b. Calculate the ΔG° at 500 K. c. Calculate the Keq values for the reaction at 298 K and 500 K. d. At what temperature value does the reaction become non-spontaneous?
Use the thermodynamic data below to determine the equilibrium constant for the conversion of oxygen to...
Use the thermodynamic data below to determine the equilibrium constant for the conversion of oxygen to ozone at 3803°C 3O2 (g) ⇌ 2O3 (g) substance       ΔH˚f (kJ/mol)     ΔG˚f (kJ/mol)      S˚ (J/mol*K)        O2 (g) 0 0 205.0        O3 (g) 142.3 163.4                        237.6
A. Using given data, calculate the change in Gibbs free energy for each of the following...
A. Using given data, calculate the change in Gibbs free energy for each of the following reactions. In each case indicate whether the reaction is spontaneous at 298K under standard conditions. 2H2O2(l)→2H2O(l)+O2(g) Gibbs free energy for H2O2(l) is -120.4kJ/mol Gibbs free energy for H2O(l) is -237.13kJ/mol B. A certain reaction has ΔH∘ = + 35.4 kJ and ΔS∘ = 85.0 J/K . Calculate ΔG∘ for the reaction at 298 K. Is the reaction spontaneous at 298K under standard conditions?
Ammonia is formed by the Haber process according to the following reaction: N2(g) + 3H2(g) ⇌...
Ammonia is formed by the Haber process according to the following reaction: N2(g) + 3H2(g) ⇌ 2NH3(g) Use the following data table to answer the questions below: Substance: ΔHf (kJ/mol) So (J/(mol*K) N2(g) 0    187.4 H2(g) 0 127.1 NH3(g) -47.3 197.6 Part 1: Using the table in the introduction, calculate the value of ΔH in units of kJ/mol. After, calculate the value of ΔS in units of J/(mol*K). Finally, cCalculate the value of ΔG in units of kJ/mol for...
1) Given the following thermochemical reaction and thermodynamic data, find Gibbs Free Energy, ΔG, and determine...
1) Given the following thermochemical reaction and thermodynamic data, find Gibbs Free Energy, ΔG, and determine if the reaction is spontaneous or non-spontaneous at 25 °C? N2(g) + 3H2(g) → 2NH3(g) ΔH = -91.8 kJ ΔS[N2] = 191 J / mol · K, ΔS[H2] = 131 J / mol · K, and ΔS[NH3] = 193 J / mol · K a.98.3 kJ; Non-Spontaneous b.-98.3 kJ; Spontaneous c.32.7 kJ; Non-Spontaneous d.ΔG = -32.7 kJ; Spontaneous 2) What is the oxidation number...
Please solve step by step for Part D Consider the Haber synthesis of gaseous NH3 (ΔH∘f=−46.1kJ/mol...
Please solve step by step for Part D Consider the Haber synthesis of gaseous NH3 (ΔH∘f=−46.1kJ/mol ;ΔG∘f=−16.5kJ/mol) : N2(g)+3H2(g)→2NH3(g) Part A Use only these data to calculate ΔH∘ and ΔS∘ for the reaction at 25 ∘C. Express your answers using three significant figures separated by a comma. ΔH∘, ΔS∘ = -92.2,-199   kJ, J/K   Part D. What are the equilibrium constants Kp and Kc for the reaction at 370 K ? Assume that ΔH∘ and ΔS∘ are independent of temperature.​ Express...
Use the thermodynamic data provided below to determine ΔG (in kJ/mol) for the condensation of C2H5OH...
Use the thermodynamic data provided below to determine ΔG (in kJ/mol) for the condensation of C2H5OH at 64.02 °C if the initial partial pressure of C2H5OH is 1.99 atm. Report your answer to one decimal place in standard notation (i.e. 123.4 kJ/mol). Substance ΔH°f (kJ/mol) S° (J mol-1K-1) C2H5OH (l) -277.7 160.7 C2H5OH (g) -235.1 282.7
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT