Concept explainers
Interpretation:
The Gibbs free energy and required change in enthalpy for a reaction system are to be calculated with given
Concept introduction:
Gibbs Free Energy can be defined as
Here,
When the change in entropy is negligible or zero, the equation changes to the expression:
Second law of thermodynamics tells about the spontaneity of a reaction and Gibbs free energy will tell in which direction the reaction is spontaneous.
It can be expressed mathematically as
Here,
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Organic Chemistry
- A reaction has AS>0 and AH>0. Which statement about this reaction must be correct? (A) If carried out in a well-insulated flask, the temperature of the reaction mixture will decrease. (B) It will occur spontancously at 298 K and 1 atm pressure. (C) As the temperature is raised, Keq for this reaction decreases. (D) The reaction has more moles of products than it has moles of reactants.arrow_forward1.4 4 (a) Calculate the value of Kc for the reaction: PCl5 (g) PCl3 (g) + Cl2 (g) ΔH = Positive Given that when 8.4 mol of PCl5 (g) is mixed with 1.8 mol of PCl3 (g) and allowed to come to equilibrium in a 10 dm3container the amount of PCl5 (g) at equilibrium is 7.2 mol. Kc =arrow_forwardFor a reaction with Keg = 0.8 and E, = 80 kJ/mol, decide which of the following statements is (are) true. Correct any false statement to make it true. Ignore entropy considerations. (a) The reaction is faster than a reaction with Keg = 8 and Ea = 80 kJ/mol. (b) The reaction is faster than a reaction with Keg = 0.8 and E, = 40 kJ/mol. (c) AG° for the reaction is a positive value. (d) The starting materials are lower in energy than the products of the reaction. (e) The reaction is exothermic.arrow_forward
- (a) Calculate the value of Kc for the reaction: PCl5 (g) PCl3 (g) + Cl2 (g) ΔH = Positive Given that when 8.4 mol of PCl5 (g) is mixed with 1.8 mol of PCl3 (g) and allowed to come to equilibrium in a 10 dm3container the amount of PCl5 (g) at equilibrium is 7.2 mol. Kc = (b) Explain the effect of the following changes below on the value of Kc: Increasing temperature Lowering the concentration of chlorine (Cl2) Addition of a catalystarrow_forward(28) The standard Gibbs free energy associated with the following reaction is -91.2 kJ: HCI (g) + NH3 (g) → NH,CI (s) What will be the value of AG,nn if the reaction takes place at 42.2°C, the partial pressure of HCI (g) is 0.788 atm and the partial pressure of NH3 (g) is 0.284 atm? (A) (B) -95.1 kl -90.7 kl (C) (D) (E) 3.93 kJ noltemol or ni lu bluos noinsou -79.5 kJ -87.3 kJarrow_forwardGiven : (a) BCl3(g) + 3H2O(l) →H3BO3(s) + 3HCl(g) ΔH = -112.5 kJ (b) B2H6(g) + 6H2O(l) →2H3BO3(s) + 6H2(g) ΔH = -493.4 kJ (c) 1/2H2(g) + 1/2Cl2(g) →HCl(g) ΔH = -92.3 kJ Calculate the value of ΔH for the reaction: B2H6(g) + 6Cl2(g) →2BCl3(g) + 6HCl(g) ΔH = ? Select one: a.-1476KJ b.-1376KJ c. -1576KJ d.-1576KJarrow_forward
- For a reaction with ΔHo = 40 kJ/mol, decide which of the following statements is (are) true. Correct any false statement to make it true. (a) The reaction is exothermic; (b) ΔGo for the reaction is positive; (c) Keq is greater than 1; (d) the bonds in the starting materials are stronger than the bonds in the product; and (e) the product is favored at equilibrium.arrow_forwardConsidering each of the following values and neglecting entropy, tell whether the starting material or product is favored at equilibrium: (a) ΔHo = 80 kJ/mol; (b) ΔHo = −40 kJ/mol.arrow_forward3. (a) Use the data given below and calculate AHO, ASO, A Gº, and Kp at 25° C for the reaction: CO (g) + 3 H₂ (g) - → CH4 (g) + H₂O (g) (b) Calculate AG for the reaction at 250 °C. (c) At what temperature (°C) is AG equal to zero? In what temperature range is this reaction product- favored? Compound CO (g) H₂(g) CH4 (g) H₂O (g) AH°, kJ/mol -110.52 0 -74.81 -241.82 So, J/mol K 197.67 130.68 186.264 188.83arrow_forward
- Question: The following reaction reaches equilibrium at the specified conditions. C6H5CH=CH2 (g) + H2 (g) <---> C6H5C2H5 (g) The system initially contains 3 mol H2 for each mole of styrene. Assume ideal gases. For styrene, ΔGof,298 = 213.18 kJ/mol, ΔHof,298 = 147.36 kJ/mol. (a) What is Ka at 600oC? (b) What are the equilibrium mole fractions at 600oC and 1 bar? (c) What are the equilibrium mole fractions at 600oC and 2 bar? Request: Can you please help me with solving part c of this problem? You can go whichever order you want, but it is part c that I need help with the most. Thank you!arrow_forwardkp and delta G ^ 0 IS NEEDED onlyarrow_forwardCalculate ΔG° for each of the following reactions from the equilibrium constant at the temperature given. (a) CO2(g) + H2(g) ⇌ CO(g) + H2O(g) t = 980°C Kp = 1.67 (b) CaCO3(s) ⇌ CaO(s) + CO2(g) t = 900°C Kp = 1.04arrow_forward
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