For the following reaction:
CS2(g) + 3O2(g) → CO2(g) +
2SO2(g)
Compound | ΔH°f (kJ mol-1) | S° (J mol-1 K-1) | ||
CS2 (g) | 116.70 | 237.80 | ||
O2 (g) | 0.00 | 205.70 | ||
CO2 (g) | -393.51 | 213.74 | ||
SO2 (g) | -296.84 | 248.20 |
Calculate ΔG°rx (in kJ) at 794.8 K for this reaction.
Report your answer to two decimal places in standard notation
(i.e. 123.45 kJ).
Assume ΔH°f and S° do not vary as a function of
temperature.
Answer:-
Firstly from the given standard value of enthalpy and entropy are used to calculate the enthalpy and entropy change for reaction then Gibbs free energy is calculated.
The answer is given in the image,
For the following reaction: CS2(g) + 3O2(g) → CO2(g) + 2SO2(g) Compound ΔH°f (kJ mol-1)...
For the following reaction: 2CH3OH(l) + 3O2(g) → 2CO2(g) + 4H2O(l) Compound ΔH°f (kJ mol-1) S° (J mol-1 K-1) CH3OH (l) -238.40 127.19 O2 (g) 0.00 205.70 CO2 (g) -393.51 213.74 H2O (l) -285.83 69.91 Determine the temperature (to two decimal places in K) such that the reaction is in equilibrium in its standard states.
For the following reaction: 2CH4(g) + O2(g) → 2CO(g) + 4H2(g) Compound ΔH°f (kJ mol-1) S° (J mol-1 K-1) CH4 (g) -74.87 188.66 O2 (g) 0.00 205.70 CO (g) -110.53 197.66 H2 (g) 0.00 130.68 Calculate ΔG°rx (in kJ) at 345.31 K for this reaction. Report your answer to two decimal places in standard notation (i.e. 123.45 kJ). Assume ΔH°f and S° do not vary as a function of temperature.
Given the following data: C(s) + O2(g) → CO2(g) ΔH = -393.5 kJ CS2(l) + 3O2(g) → CO2(g) + 2SO2(g) ΔH = -1076.5 kJ C(s) + 2S(s) → CS2(l) ΔH = +89.4 kJ Find the ΔH of the following reaction: SO2(g) → S(s) + O2(g)
Given: C(s) + O2(g) ---> CO2(g) ΔH = −393.5 kJ/mol S(s) + O2(g) ---> SO2(g) ΔH = −296.8 kJ/mol C(s) + 2S(s) ---> CS2(ℓ) ΔH = +87.9 kJ/mol A) Calculate the standard enthalpy change for the following reaction CS2(ℓ) + 3O2(g) ---> CO2(g) + 2SO2(g) ΔH° rxn = -1075 kJ/mol B) Using the equation and standard enthalpy change for the reaction (from part A), calculate the amount of heat produced or consumed when 3.2 mol of CS2 reacts with excess...
For the following reaction: 2C3H18(1) + 2502(g) → 16C02(g) + 18H20(1) Compound AH°F (kJ moll) Sº (3 mol-1 K-1) C8H18 ( 259.30 328.00 O2 (g) 0.00 205.70 CO2 (9) -393.51 213.74 H20 (1) -285.83 69.91 Determine the temperature (to two decimal places in K) such that the reaction is in equilibrium in its standard states.
For the following reaction: 2Na(s) + 2H2O(l) → 2NaOH(s) + H2(g) Compound ΔH°f (kJ mol-1) S° (J mol-1 K-1) Na (s) 0.00 51.30 H2O (l) -285.83 69.91 NaOH (s) -425.93 64.46 H2 (g) 0.00 130.68 Calculate ΔG°rx (in kJ) at 391.96 K for this reaction. Assume ΔH°f and S° do not vary as a function of temperature.
Use the given data at 500 K to calculate ΔG°for the reaction 2H2S(g) + 3O2(g) → 2H2O(g) + 2SO2(g) Substance H2S(g) O2(g) H2O(g) SO2(g) ΔH°f(kJ/mol) -21 0 -242 -296.8 S°(J/K·mol) 206 205 189 248
Consider the following reaction. CS2(g) + 3O2(g) CO2(g) + 2SO2(g) If the volumes of all products and reactants are measured at 75°C and 1 atm, what is the total volume of products if 5.00 L of CS2 and 18.0 L of O2 are used?
Find the enthalpy change for the reaction CS2(l) + 3 O2(g) → CO2(g) + 2 SO2(g) when: C(s) + O2(g) → CO2(g) ΔHf = -393.5 kJ/mol S(s) + O2(g) → SO2(g) ΔHf = -296.8 kJ/mol C(s) + 2 S(s) → CS2(l) ΔHf = 87.9 kJ/mol
If the ΔH°rx and ΔS°rx for a gas phase chemical reaction are 419.53 kJ and -178.85 J/K respectively, determine ΔG°rx (in kJ) at 103.11 °C. Report your answer to two decimal places (i.e. 123.45 kJ). Assume that ΔH°rx and ΔS°rx do not vary with temperature.