HERMODYNAMICS: Given the following hypothetical thermochemical equations:2QJ3 (g) + 3Q2X (g) → 4Q2 (g) + 3J2X (I)AH = -1022.3 kJ4QJ3 (g) + 3X2 (g) → 2Q2 (g) + 6J2X (I)AH = -1517.2 kJQ2 (g) + ½ X2 (g)→ Q2X (g)Calculate the value of AH (in kJ) for the reaction:(Round off final answer to ONE decimal place)

Answered: Image /qna-images/answer/28898bae-5c05-45ce-a550-2e7cb82f7e5b.jpg

HERMODYNAMICS: Given the following hypothetical thermochemical equations: 2QJ3 (g) + 3Q2X (g) → 4Q2 (g) + 3J2X (I) AH = -1022.3 kJ 4QJ3 (g) + 3X2 (g) 2Q2 (g) + 6J2X (1) AH = -1517.2 kJ Calculate the value of AH (in kJ) for the reaction: Q2 (g) + ½ X2 (g) Q2X (g) (Round off final answer to ONE decimal place)

HERMODYNAMICS: Given the following hypothetical thermochemical equations: 2QJ3 (g) + 3Q2X (g) → 4Q2 (g) + 3J2X (I) AH = -1022.3 kJ 4QJ3 (g) + 3X2 (g) 2Q2 (g) + 6J2X (1) AH = -1517.2 kJ Calculate the value of AH (in kJ) for the reaction: Q2 (g) + ½ X2 (g) Q2X (g) (Round off final answer to ONE decimal place)

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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When 1.16E-1 g of Zn(s) combines with 5.21E1 mL of HCl(aq) in a coffee cup calorimeter, all of the zinc reacts, which increases the temperature of the HCl solution from 2.34E1 °C to 2.46E1 °C:Zn(s) + 2HCl(aq) → ZnCl(aq) + H(g) Calculate the enthalpy change of the reaction ΔH in J/mol. (Assume the volume of the solution doesn't change, density of the solution is 1.00 g/mL and the specific heat capacity of solution is 4.184 J/g°C.)
Consider these reactions: Reaction 1: H2(g) + Cl2(g) 2HCI(g) AH = -184.6 kJ Reaction 2: 20F2(g)- O2(g) + 2 F2(g) AH = -49.4 kJ Reaction 3: N2(g) + 202(g) 2NO2(g) AH = +66.4 kJ Reaction 2 can be described as an [Select] [ Select ] reaction
THERMODYNAMICS: Given the following hypothetical thermochemical equations: 2QJ3 (g) + 3Q2X (g) – 4Q2 (g) + 3J2X (1) AH = -1,070 kJ 4QJ3 (g) + 3X2 (g) → 202 (g) + 6J2X (1) AH = -1,580.2 kJ Calculate the value of AH (in kJ) for the reaction: Q2 (g) + % X2 (g) –→ Q2X (g) (Round off the final answer to ONE decimal place. Do not include unit.)
2. Consider the following thermochemical equations: C(graphite) + O2 (g) → co2 (g) AH°rn = -393.5 kJ H2 (g) + % 02 (g) → H;0 (/) AHO = -285.8 kJ 2 C:Hs (g) + 702 (g) → 4 CO2 (g) + 6 H20 (1) AH°rn = -3119.6 kJ a. Derive the thermochemical equation describing the combustion of two moles of C(graphite). b. Derive the thermochemical equation describing the combustion of three moles of H2 (g). c. Derive the thermochemical equation describing the synthesis of one mole of C,Hg (9) from carbon dioxide and water. d. Use Hess's Law and the three thermochemical equations you derived in parts a-c to determine the AH°ran for the following reaction: 2C (graphite) + 3 H2 (g) → C;Hs (g)
THERMODYNAMICS: Given the following hypothetical thermochemical equations: 2QJ3 (g) + 3Q2X (g) – 4Q2 (9) + 3J2X (1) AH = -1,017.3 kJ 4QJ3 (g) + 3X2 (g) → 202 (9) + 6J2X (1) AH = -1,548.8 kJ Calculate the value of AH (in kJ) for the reaction: Q2 (g) + % X2 (g) –→ Q2X (g) (Round off the final answer to ONE decimal place. Do not include unit.)
THERMODYNAMICS: Given the following hypothetical thermochemical equations: 2QJ3 (g) + 3Q2X (g) – 4Q2 (g) + 3J2X (1) AH = -1,096.4 kJ 4QJ3 (g) + 3X2 (g) → 202 (g) + 6J2X (1) AH = -1,548.8 kJ Calculate the value of AH (in kJ) for the reaction: Q2 (g) + % X2 (g) –→ Q2X (g) (Round off the final answer to ONE decimal place. Do not include unit,.)
THERMODYNAMICS: Calculate the value of AH (in kJ) for the reaction, 2AR3 (9) + Q (s) → A2R4 (g) + QR2 (s) given the following hypothetical thermochemical equations: 2AR3 (g) + 2AX (g) → A2R4 (9) + 2XAR (g) ΔΗ = -88.4 kJ AX (g) + ½R2 (g) → XAR (g) AH = -164.8 kJ Q (s) + R2 (g) –→ QR2 (s) AH = -520.4 kJ (Round off the final answer to ONE decimal place. Do not include the unit.)
Given the following thermochemical equations: C2H29) + 5 02(9) 2C02(g) + H201) AH° = -1229.5 kJ AH° = -393.5 kJ C(6) + 02(g) → C02(g) 1 H2(9) +5 02(9) → H20) AH° = -285.5 kJ 2 Calculate the enthalpy (in kJ) for this reaction: 206) + H2(g) → C,H2(9)
Rearrange the thermochemical equations given to determine the AH value for the equation of interest. Determine: 2 Fe203(s) + 3 C(s) → 3 CO2(g) + 4 Fe(s) Given Enthalpy (K/mol) 3 -833.8 O2(g) Fe203(s) 2 1. 2 Fe(s) + ch 2. C(s) + - O2(8) → CO(g) -110.4 1 O2(g) → CO2(g) -264.8 3. CO(g) + K/mol
THERMODYNAMICS: Calculate the value of AH (in kJ) for the reaction, 2AR3 (g) + Q (s)→ A2R4 (g) + QR2 (s) given the following hypothetical thermochemical equations: 2AR3 (g) + 2AX (g) → A2R4 (g) + 2XAR (g) AH = -87.2 kJ AX (g) + ½R2 (g) → XAR (g) AH = -157.3 kJ Q (s) + R2 (g) – QR2 (s) AH = -510.2 kJ (Round off the final answer to ONE decimal place. Do not include the unit.)
Use the thermochemical equations shown below to determine the enthalpy for the reaction: C3H8(g) + 502(g) +3CO 2 + 4H2O(1) CO2 C(graphite) + 02 HE = 221.6KJ H2(g) + 1/2O2(g) H2O(1) HE= -160.3KJ 3 C(graphite) + 4 H2(g) C3H8(g) HE= -58.5KJ
Question 4 THERMODYNAMICS: Calculate the value of AH (in kJ) for the reaction, 2AR3 (g) + Q (s) → A2R4 (g) + QR2 (s) given the following hypothetical thermochemical equations: 2AR3 (g) + 2AX (g) → A2R4 (g) + 2XAR (g) AH = -81.9 kJ AX (g) + ½R2 (g) → XAR (g) AH = -159.1 kJ Q (s) + R2 (g) → QR2 (s) (Round off final answer to ONE decimal place) AH = -510.4 kJ Round your answer to 1 decimal place.