### Entropy and Free Energy: Calculating ΔG from ΔH and ΔSA chemical engineer is studying the two reactions shown in the table below.In each case, he fills a reaction vessel with some mixture of the reactants and products at a constant temperature of 143.0 °C and constant total pressure. Then, he measures the reaction enthalpy ΔH and reaction entropy ΔS of the first reaction, and the reaction enthalpy ΔH and reaction free energy ΔG of the second reaction. The results of his measurements are shown in the table.Complete the table. That is, calculate ΔG for the first reaction and ΔS for the second. (Round your answer to zero decimal places.) Then, decide whether, under the conditions the engineer has set up, the reaction is spontaneous, the reverse reaction is spontaneous, or neither forward nor reverse reaction is spontaneous because the system is at equilibrium.#### Results Table:| Reaction | ΔH (kJ) | ΔS (J/K) | ΔG (kJ) | Which is spontaneous? ||----------|---------|----------|---------|-----------------------|| P₄(s) + 6Cl₂(g) → 4PCl₃(g) | -1207 | -2911 | **[Calculate]** | [This reaction / Reverse reaction / Neither] || 6C(s) + 6H₂(g) + 3O₂(g) → C₆H₁₂O₆(s) | -1237 | **[Calculate]** | 0 | [This reaction / Reverse reaction / Neither] |#### Instructions for Completing the Table:1. **Calculate ΔG for the First Reaction:** Use the formula ΔG = ΔH - TΔS, where T is the temperature in Kelvin. Convert the given temperature to Kelvin by adding 273.15 to the Celsius temperature.2. **Calculate ΔS for the Second Reaction:** Use the provided relationship and data given in the table.3. **Determine Spontaneity:** After calculating ΔG, determine if the reaction is spontaneous by evaluating the sign of ΔG: - If ΔG < 0, the reaction is spontaneous. - If ΔG > 0, the reverse reaction is spontaneous. - If ΔG = 0, neither forward nor reverse reaction is spontaneous (the

∆G = ∆H - T∆SWhere,∆G = gibbs free energy∆H = reaction enthalpyT = temperature∆S = reaction entropy•…

P₁ (g) + 6C1₂ (g) 4PC1, (g) 6C(s) + 6H₂(g) + 30₂(g) → CH₁₂06(s) ΔΗ -1207. KJ AS = -2911. AG - KJ Which is spontaneous? O this reaction ΔΗ J - K AS= AG the reverse reaction -0--/- neither -1237. KJ 0. kJ Which is spontaneous? O this reaction the reverse reaction O neither X S

P₁ (g) + 6C1₂ (g) 4PC1, (g) 6C(s) + 6H₂(g) + 30₂(g) → CH₁₂06(s) ΔΗ -1207. KJ AS = -2911. AG - KJ Which is spontaneous? O this reaction ΔΗ J - K AS= AG the reverse reaction -0--/- neither -1237. KJ 0. kJ Which is spontaneous? O this reaction the reverse reaction O neither X S

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Use the table to calculate AG" for the following reactions at 25°C. rxn Species AG° f (kJ/mol) Mg(s) |02g) MgO(s) -569.6 SO2(g) -300.57 SO3(g) |-370.53 C2H2(g) 209.2 |CO2(g) -394.65 H20(1) -237.35 H20(g) -228.77 (a) 2 Mg(s) + O2g) → 2 MgO(s) kJ (b) 2 SO2(g) + O2g) → 2 SO3(g) kJ (c) 2 C2H2(g) + 5 02(g) → 4 CO2(g) + 2 H20(!) kJ
21. Given the following two equations CH5OH(1) + 3 O2(8) 2 СO2(g) + 3 Н20(1) AH = -1370 kJ С-Н(8) + 3 О2(s) → 2 CO2(g) + 2 H2O(1) AH = -1410 kJ Calculate AH for the reaction C2H4(8) + H2O(1) CH5OH(I) ΔΗ- a. -2780 kJ b. 2780 kJ с. 40 kJ d. 40 kJ
For the reaction |C(s,graphite) + O(g) CO2(g) |AG° = -394.4 kJ and AS° = 2.9 J/K at 315 K and 1 atm. This reaction is (reactant, product) | favored under standard conditions at 315 K. The standard enthalpy change for the reaction of 2.03 moles of C(s,graphite) at this temperature would be kJ.
Given: H2(g) + 1/2 O2 (g) —> H2O (l) Delta H = -285.8 kJ Combine this equation with equation 1 and 2, to obtain the equation for the combustion of magnesium Equation 1: MgO + 2HCl —> MgCl2 + H2O Equation 2: Mg(s) + 2 HCl (aq) —> MgCl2 (aq) + H2 (g)
Gaseous SO2Cl2 reacts with liquid water to form hydrogen chloride gas and liquid sulfuric acid. Using data tabulated in Appendix G of your text, calculate the enthalpy change, in units of kJ/mol, for the formation of 2 mol of liquid sulfuric acid. The answer is no -139.
6C(s) + 6H₂(g) + 30₂(g) → CH₁₂O6 (s) 2Al(s) + Fe₂O₂ (s) → ALO₂ (s) + 2Fe(s) AH = - 1237. kJ AS = -3280. AG = kJ Which is spontaneous? this reaction O the reverse reaction neither AH = -852. kJ AS = J K J K AG 19. kJ Which is spontaneous? this reaction the reverse reaction neither
A chemical engineer is studying the two reactions shown in the table below. In each case, he fills a reaction vessel with some mixture of the reactants and products at a constant temperature of 15.0 °C and constant total pressure. Then, he measures the reaction enthalpy AH and reaction entropy AS of the first reaction, and the reaction enthalpy AH and reaction free energy AG of the second reaction. The results of his measurements are shown in the table. Complete the table. That is, calculate AG for the first reaction and AS for the second. (Round your answer to zero decimal places.) Then, decide whether, under the conditions the engineer has set up, the reaction is spontaneous, the reverse reaction is spontaneous, or neither forward nor reverse reaction is spontaneous because the system is at equilibrium. 6C(s) + 6H₂(g) + 30₂(g) 2Al(s) + 3H₂O(g) C6H₁2O6 (s) Al₂O3 (s) + 3H₂(g) AH = -1237. kJ AS-4214. AG = KJ Which is spontaneous? this reaction the reverse reaction neither ΔΗ = -951. kJ…
ENTROPY AND FREE ENERGY Using the general properties of Gibbs free energy The standard reaction free energy AG = -910. kJ for this reaction: 6C(s) + 6H₂(9) + 30₂(9)-CH12O6(s) Use this information to complete the table below. Round each of your answers to the nearest kJ. reaction 12C(s) + 12H₂(g) + 60₂(g) → 2C,H,O,(s) - C₂H₁20 (9) 3C(s) + 3H₂(g) + 0₂ (4) CH₁20,(s) 6C(s) + 6H₂(g) + 30₂ (9) 1 AGⓇ kJ kJ 0 kJ 00
If AS at 25°C sys =-198.3 J/K, is the reaction spontaneous N2 + 3H2 → 2NH3 • AS univ = AS sys + AS surr > ? AS surr (AHsys)/T Determine the AHrxn • AH° N2(g) = 0 kJ/mol • AH° H2(g) = 0 kJ/mol • AH°; NH3(g) = -46.19 kJ/mol
Classify each of the reactions according to one of the four reaction types summarized in the table below. Reaction Spontaneous Process? Туре ΔΗ' (system) AS (system) (Standard Conditions) 1 Exothermic, 0 Spontaneous at all temperatures. AS (universe) > 0. Exothermic, 0 Positive, > 0 Depends on relative magnitudes of AH and AS". Spontaneous at higher temperatures. 3 Endothermic, > 0 Negative, < o Not spontaneous at any temperature. AS° (universe) < 0. 4 C6 H12O6 (s) + 6 02 (g) → 6 CO2 (g) + 6 H2O(l) a. A,H° = -673 kJ/mol-rxn A,S° = 60.4 J/K · mol-rxn O Spontaneous at all temperatures. O Spontaneous at lower temperatures. O Spontaneous at higher temperatures. ONot spontaneous at any temperature. b. ВаО(s) + C(graphite) — Ba(s) + СО(g) A,H° = 437.6 kJ/mol-rxn A,S° = 184.9 J/K· mol-rxn OSpontaneous at all temperatures. O Spontaneous at lower temperatures. O Spontaneous at higher temperatures. ONot spontaneous at any temperature.
From the enthalpies of reaction 2C(s) + O2 (9) → 2C0(g) 2C(s) + O2 (9) + 4H2 (g) → 2CH3 OH(g) AH ΔΗ :-221.0kJ %3| :-402.4kJ calculate AH for the reaction CO(g) + 2H2 (g) → CH3 OH(g) Express your answer to three significant figures with the appropriate units.
If AS at 25°C sys =-198.3 J/K, is the reaction spontaneous N2 + 3H2 → 2NH3 • AS univ = AS sys + AS surr > ? AS surr (AHsys)/T Determine the AHrxn • AH° N2(g) = 0 kJ/mol • AH° H2(g) = 0 kJ/mol • AH°; NH3(g) = -46.19 kJ/mol