A series of hypothetical thermochemical reactions with their corresponding enthalpies aregiven below.As) + X2(9) → AX2(9)A;Q8(g) + 5X2(9) → 3AX2(9)Q2(9)AH = - 394 KJ/mole+ 4Q2X) AH = - 2,222 KJ/mole+ ½ X2lg) → QzXAH = - 286 KJ/mole(A). Using the thermochemical equations above, the heat for the given overall reaction isKilojoules per mole. Express answer in the NEAREST WHOLE NUMBER+ 4Q2(9) > A,Q3(9)3A(s)(B). If the overall thermochemical reaction given above has an entropy value of -270Joules/Kelvin, its free energy (AG) at 100 degrees Celsius isin the NEAREST WHOLE NUMBER(C). If the overall thermochemical reaction given above has an entropy value of -270Joules/Kelvin, then it will become spontaneous at temperatures aboveKiloJoules. Express answerKelvin. Expressanswer in the NEAREST WHOLE NUMBER

A series of hypothetical thermochemical reactions with their corresponding enthalpies are given below. As) + X2(9) → AX2(g) A,Qa(9) + 5X2(g) + ½ X2(g) > AH = - 394 KJ/mole > ЗАХ, → 3AX2(9) + 4Q,Xm AH = - 2,222 KJ/mole Q2(9) AH = - 286 KJ/mole (A). Using the thermochemical equations above, the heat for the given overall reaction is Kilojoules per mole. Express answer in the NEAREST WHOLE NUMBER 3A(s) + 4Q2(g) → A;Q8(g) A,Qs(9) (B). If the overall thermochemical reaction given above has an entropy value of -270 Joules/Kelvin, its free energy (AG) at 100 degrees Celsius is_ KiloJoules. Express answer in the NEAREST WHOLE NUMBER (C). If the overall thermochemical reaction given above has an entropy value of -270 Joules/Kelvin, then it will become spontaneous at temperatures above_ Kelvin. Express answer in the NEAREST WHOLE NUMBER

A series of hypothetical thermochemical reactions with their corresponding enthalpies are given below. As) + X2(9) → AX2(g) A,Qa(9) + 5X2(g) + ½ X2(g) > AH = - 394 KJ/mole > ЗАХ, → 3AX2(9) + 4Q,Xm AH = - 2,222 KJ/mole Q2(9) AH = - 286 KJ/mole (A). Using the thermochemical equations above, the heat for the given overall reaction is Kilojoules per mole. Express answer in the NEAREST WHOLE NUMBER 3A(s) + 4Q2(g) → A;Q8(g) A,Qs(9) (B). If the overall thermochemical reaction given above has an entropy value of -270 Joules/Kelvin, its free energy (AG) at 100 degrees Celsius is_ KiloJoules. Express answer in the NEAREST WHOLE NUMBER (C). If the overall thermochemical reaction given above has an entropy value of -270 Joules/Kelvin, then it will become spontaneous at temperatures above_ Kelvin. Express answer in the NEAREST WHOLE NUMBER

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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4 CH3NH2 (g) + 2 H2O (l) ➝ CO2 (g) + 3 CH4 (g) + 4 NH3 (g) ΔHº = –138.96 kJ• Calculate the overall change in enthalpy when 30.00 grams of CH3NH2 (g) reacts with excess H2O (l). [Be sure to include the correct sign.]
Given the two thermochemical equations below, find the enthalpy change for the reaction: N2H4 (l) + H2 (g) → 2 NH3 (g) delta H = ? Given: N2O4 (g) + 3 H2 (g) → 2 NH3 (g) + 2 O2 (g) delta H = – 100.96 kJ N2O4 (g) + 2 H2 (g) → N2H4 (l) + 2 O2 (g) delta H = + 41.47 kJ Select one: a. – 142.43 kJ b. – 59.49 kJ c. + 41.47 kJ d. + 59.49 kJ e. + 142.43 kJ f. none of these g. – 100.96 kJ
4. Hess's Law Calculate AH for the reaction: 2 C(s) + H2(g) → C:H2(g), given the following chemical equations and their respective enthalpy changes: C2H2(g)+02(g)→2CO2(g)+H20(1) AH, = -1299.6 kJ (1) %3D C(s)+02(g)→CO2(g) AH, = -393.5 kJ (2) H2(g)+502(g)→H20(1) AH3 = -285.8 kJ (3)
For which one of the following is the enthalpy of the reaction the same as the enthalpy of formation? A) 2 H (g) + Se (s) → H:Se (g) B) KOH (s) → K*(aq) + OH- (aq) C) 2 Fe (s) + 3 Cl2 (g) → 2 FeCl3 (s) D) Ni (s) + ½ O2 (g)- NiO (s)
Given: 4AICI3(s) + 302(g) → 2Al,O3(s) + 6C12(g); AH=-529.0 kJ determine AH for the following thermochemical equation. Cl,(3) + Al,O;(s) → AICI,(9) + 02(g) O +88.2 kJ O +264.5 kJ O +176.3 kJ O -176.3 kJ O +529.0 kJ
Use Hess’s law to calculate the enthalpy change for the reaction WO3(s) + 3H2(g) —> W(s) +3H2O(g) from the following data: 2W(s) + 3O2(g) —> 2 WO3(s) H=-1685.4 2H2(g) + O2(g) —> 2H2O(g) H=-477.84
What is the standard enthalpy change for the reaction: 2C5H10(1) + 15O2(g) → 10CO2(g) + 10H20(1); where C5H10(1) AH°f = –605.0 kJ/mol; CO2(g) AH°f = -393.5 kJ/mol; H2O(1) AH°F - 285.8 kJ/mol. O a. -5583 kJ O b. -5783 kJ O C. -5383 kJ O d. -5183 kJ e. -4983 kJ
Find the standard change in enthalpy for the following reaction. N₂H4(1) + O₂(g) → 2 NO(g) + 2 H₂O(g) Given the following known reactions: N₂(g) + 2 H₂(g) → N₂H₁ AH = 97.4 kJ/mol N₂(g) + O₂(g) → NO(g) AH = 90.2 kJ/mol H₂(g) + O₂(g) → H₂O(g) AH = -242 kJ/mol Remember you need to include O₂ to get a balanced reaction but its AH = 0 kJ/mol !!
（1）Consider the reaction: 2A (g) + 3 B (g) → 2 C (g) ΔHrxn = +254.3 kJ What will be the enthalpy change (in kJ) if 0.471 mol B reacts in excess A? （2）Consider the reaction: C (s) + O2 (g) → CO2 (g) ΔHrxn = -393.5 kJ What mass of carbon (in g) must be reacted via this mechanism to release 581.2 kJ of heat?
You wish to know the enthalpy change for the formation of liquid PC13 from the elements. P4(s) + 6 Cl2 (g) → 4 PC13 (€) A, H' =? The enthalpy change for the formation of PCI5 from the elements can be determined experimentally, as can the enthalpy change for the reaction of PCI3 (() with more chlorine to give PCI5 (s): P4(s) + 10 Cl2 (g) → 4 PCI5 (s) A,H = -1774.0 kJ/mol – rxn PCI3 (e) + Cl2 (g) → PC15 (s) A,H° = -123.8 kJ/mol – rxn Use these data to calculate the enthalpy change for the formation of 3.60 mol of PCI3 (€) from phosphorus and chlorine. Enthalpy change = | kJ