Consider the reaction:\[ 6\text{CO}_2 (g) + 6\text{H}_2\text{O} (l) \rightarrow \text{C}_6\text{H}_{12}\text{O}_6 (s) + 6\text{O}_2 (g) \]Using standard thermodynamic data at 298 K, calculate the entropy change for the surroundings when 1.87 moles of \(\text{CO}_2 (g)\) react at standard conditions.### Thermodynamic Data Table| Substance | \(\Delta H_f^\circ\) (kJ/mol) ||-----------|-------------------------------|| \(\text{CO}_2 (g)\) | –393.5 || \(\text{H}_2\text{O} (l)\) | –285.8 || \(\text{C}_6\text{H}_{12}\text{O}_6 (s)\) | –1275.0 || \(\text{O}_2 (g)\) | 0.0 |### Calculating Entropy Change\[ \Delta S^\circ_{\text{surroundings}} = \underline{\hspace{1cm}} \, \text{J/K} \]To calculate the entropy change for the surroundings, remember to use the formula:\[ \Delta S_{\text{surroundings}} = - \frac{\Delta H_{\text{reaction}}}{T} \]Where:- \(\Delta H_{\text{reaction}}\) is the enthalpy change for the reaction,- \(T\) is the temperature in Kelvin (298 K in this case).

To calculate the deltaS°surr , first we will calculate deltaH°rxn for given reaction . Then we will…

Consider the reaction: 6CO2 (g) + 6H₂O(l) → C6H12O6 + 602(g) Using standard thermodynamic data at 298 K, calculate the entropy change for the surroundings when 1.87 moles of CO₂ (g) react at standard conditions. ASO Substance AH (kJ/mol) CO₂(g) -393.5 H₂O(1) -285.8 C6H12O6 -1275.0 O2(g) 0.0 surroundings = J/K

Consider the reaction: 6CO2 (g) + 6H₂O(l) → C6H12O6 + 602(g) Using standard thermodynamic data at 298 K, calculate the entropy change for the surroundings when 1.87 moles of CO₂ (g) react at standard conditions. ASO Substance AH (kJ/mol) CO₂(g) -393.5 H₂O(1) -285.8 C6H12O6 -1275.0 O2(g) 0.0 surroundings = J/K

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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i need help with these pleasee Consider the reaction: 2NH3(g) +202(g) → N₂O(g) + 3H₂O(1) Using standard thermodynamic data at 298 K, calculate the entropy change for the surroundings when 1.64 moles of NH3(g) react at standard conditions. ΔS Substance AH (kJ/mol) NH3(g) As O₂(g) N₂O(g) H₂O (1) surroundings Consider the reaction: system -46.1 0.0 82.1 -285.8 H₂CO(g) + O2(g) → CO₂(g) + H₂O(1) Using standard absolute entropies at 298 K, calculate the entropy change for the system when 1.63 moles of H₂CO(g) react at standard conditions. J/K J/K Substance Sº (J/K mol) H₂CO(g) 219.0 0₂ (9) 205.1 CO₂(g) 213.7 H₂O(l) 69.9 SUVIL
Consider the reaction: 4HCl(g) + O₂(g) → 2H₂O(g) + 2Cl₂ (g) Using standard absolute entropies at 298 K, calculate the entropy change for the system when 1.69 moles of HCl(g) react at standard conditions. AS⁰ Substance So (J/K mol) HCl(g) 186.9 O2(g) 205.1 H₂O(g) 188.8 Cl₂ (g) 223.1 system = J/K
The energy produced by a spontaneous reaction is often used to provide the energy to drive nonspontaneous reactions. For example, the conversion of ATP to ADP in our cells acts as the energy source for the nonspontaneous reactions that combine amino acids during protein synthesis. Here is the conversion of ATP to ADP: ATP(aq) + H,O = ADP(aq) + HPO,²-(aq) AG° = -30.0 kj/mol A typical reaction combining two amino acids during protein synthesis has a AG = 15.0 kJ/mol. What is the minimum ATP concentration required to provide the energy to combine 1 mole of amino acid synthesis (total energy = 15.0 kJ)? Assume that the concentration of ADP = 9 µM (micromolar) and HPO42- = 4 mM (millimolar). %3! %3D
For the following reaction (the detonation of nitroglycerin), is it spontaneous under standard conditions and how can you tell? 4 C4H5N3O9(l) (arrow >) 12 CO2(g) + 10 H2O(g) + 6 N2(g) + O2(g) ΔH = -5678 kJ
Consider the reaction: NH4Cl(aq) →→→→NH3(g) + HCl(aq) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 1.90 moles of NH4Cl(aq) react at standard conditions. AS° surroundings = J/K
Consider the reaction HCI(g) + NH3(g)- →NH4Cl(s) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 2.34 moles of HCI(g) react at standard conditions. ASO surroundings = J/K
AG (ky/mol) SUbStance / 1on so (ky/mol) -1206.9 /moi k) 92.9 • CaC03l>) -|128.8 • CaO (s) |- 635. I – 60 3.5 38.2 • CO2(9) - 393.5 - 394.4 213.7 191.5 3 H2(g) 202.7 ZNH3 ) -45.9 -16 193 • Hz (g) 202.7 223.0 -92.31 - 95. 30 186.79
For the reactionNH4NO3(aq) N2O(g) + 2 H2O(l)G° = -175.3 kJ and H° = -149.6 kJ at 257 K and 1 atm.This reaction is (reactant, product) favored under standard conditions at 257 K.The entropy change for the reaction of 1.74 moles of NH4NO3(aq) at this temperature would be J/K.
Enthalpy and Gibb's Free Energy Chemical energy is released or absorbed from reactions in various forms. The most easily measurable form of energy comes in the form of heat, or enthalpy. The enthalpy of a reaction can be calculated from the heats of formation of the substances involved in the reaction: AHxn = AH₂ (products) - AH (reactants) Entropy change, AS°, is a measure of the number of energetically equivalent microstates introduced into the system during the reaction. The degree of spontaneity of a reaction is represented by the Gibbs free energy, AGO. The Gibbs free energy depends on both the enthalpy and entropy changes that take place during the reaction: AG=AH° - TAS° where T is standard temperature, 298 K. ▼ Part A Calculate the standard enthalpy change for the reaction where the heats of formation are given in the following table: ΔΗ Substance (kJ/mol) A B C D -241 -407 191 -501 Express your answer in kilojoules. ► View Available Hint(s) {—| ΑΣΦ AHixn= 2A+B=2C+2D ? kJ…