The enthalpy changes for three compounds are shown below: N2(g) + 5/2 O2(g) + 15.0 kJ H2(g) + 1/2 02(g) 1/2 H2(g) + 1/2 N2(g) + 3/2 02(g) N₂O5(8) H₂O(1) + 285.8 kJ A/ HNO3(1) + 174.1 kJ Net: N₂O5(g) + H₂O (1) 2 HNO3(1) The standard enthalpy change for the net reaction is: ab.c kJ Input the value for ab.c in the blank below. Include a positive or negative sign in your answer: eg -16.8 or +16.8 • Do not include units Is the reaction endothermic or exothermic? • write exothermic or endothermic in the blank

The enthalpy changes for three compounds are shown below: N2(g) + 5/2 O2(g) + 15.0 kJ H2(g) + 1/2 02(g) 1/2 H2(g) + 1/2 N2(g) + 3/2 02(g) N₂O5(8) H₂O(1) + 285.8 kJ A/ HNO3(1) + 174.1 kJ Net: N₂O5(g) + H₂O (1) 2 HNO3(1) The standard enthalpy change for the net reaction is: ab.c kJ Input the value for ab.c in the blank below. Include a positive or negative sign in your answer: eg -16.8 or +16.8 • Do not include units Is the reaction endothermic or exothermic? • write exothermic or endothermic in the blank

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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Coffee cup calorimetry was used to determine the enthalpy of solution of lithium iodide (LiI): A) First, the coffee cup was calibrated. 50.0 mL of water (s=4.184J/g×0 C) was heated to 95.50C and mixed with 50.0 mL of cool water at 22.30 C. The temperature of the contents of the coffee cup stabilized at 49.40 C. Determine the heat capacity of the coffee cup. B) Next, the enthalpy of solution of lithium iodide (LiI) was determined. 0.0g of solid lithium iodide was added to 100mL of water at 23.20 C in the coffee cup, and the temperature increased to 32.10 C. Calculate the enthalpy of the solution of lithium iodide.
Q2 (a) Given that the enthalpy values for the following reactions are: AH = -1049.0 kJ 2 Al (5) + 6HCI (ад) — 2 ALCI3 (аg) + 3 Н2 (9) AH = -74.8 kJ HCl (g) → HCl (aq) + Cl2 (g) → 2HCI AH = -185.0 kJ H2 (9) + AlCl3 (s) → AlCl3 (aq) (g) AH = -323.0 kJ ΔΗ Manipulate the reactions above to determine AH for: 2 Al(s) + 3 Cl2 (9) → 2 AlCl3 (s) In one experiment, a 25 g sample of an alloy was heated to a certain temperature. (b) The heated alloy was then dropped into a beaker containing 90 g of water at 25.32 °C. The temperature of the water rose to a final value of 27.18 °C. Predict the final temperature that the alloy is being heated at the beginning of the experiment. Given that the heat capacity of water and alloy are 4.184 Jgl°C-' and 0.3847 Jg l°C' respectively.
Calculate the enthalpy change for the following reaction, _C₂H4 (9) + using the following standard heats of formation: AH(C₂H4) = +52.5 kJ/mol; AH(CF4) =-933.0 kJ/mol; AH (HF) =-271.6 kJ/mol _F₂(g) → _CF4(g) + _HF(g)
Given the enthalpies for the following reactions: 2 Fe (s) + 3/2 O₂(g) → Fe2O3 (s)Math output error AH₂ ΔΗ, CO (g) + 1/2 O₂ (g) → CO₂ (g) CO2 (g) ΔΗ = -248.7 kJ rxn rxn = 1,091.6 kJ Calculate the enthalpy change (AHrxn) for the following reaction: Fe2O3 (s) + 3 CO (g) → 2 Fe (s) + 3 CO₂ (g)
Use standard enthalpies to determine the enthalpy change from the following reaction occurring: (1:3) 2Cl₂(g) + 2H₂O(g) →4HCl(g) + O₂(g)
and 2 Enthalpy #1.30 enoindice pled TT 2010 2D49.bs 22) When 1 mole of Fe2O3(s) reacts with H₂(g) to form Fe(s) and H2O(g) according to the following equation, 98.8 kJ of energy are absorbed. (6 Fe2O3(s) + 3 H2(g) → 2 Fe(s) + 3 H₂O(g) (pp) (20/d4 (pm) $40 D Reaclanis 13:48 SH Products endothermic, A endothermic, B (2) exothermic, A exothermic, B nad W (21 Enthalpy MOMS. Reactants Products (A) (B) Is the reaction endothermic or exothermic, and which of the enthalpy diagrams above represents grow olor and to noin (01 this reaction? SH GH TH A HOMH X HILD 08.10 Page 6 of 8
1.Given reactions A - C, determine the the change in enthalpy (kJ) for the reaction between C(s) + H2(g). Do not type units into your answer. C(graphite) + 2 H2(g) ⟶⟶ CH4(g) Δ?=?ΔH=? A. C(graphite) + O2(g) ⟶⟶ CO2(g) Δ?=−393.51??ΔH=−393.51kJB. H2(g) + 1/2 O2(g) ⟶⟶ H2O(l) Δ?=−285.83??ΔH=−285.83kJC. CO2(g) + 2 H2O(l) ⟶⟶ CH4(g) + 2 O2(g) Δ?=890.30??ΔH=890.30kJ 2. Based on your answer to the previous problem, determine the value of q (kJ) if 5.08 grams of methane (CH4) gas is produced from graphite and hydrogen gas. Do not type units into your answer.
Help with number 3 please
Use Hess’s law to obtain the enthalpy change for this reaction: H2(g) + 1/8 S8(rhombic) -> H2S(g) from the following enthalpies:H2S(g) + 3/2 O2(g) à H2O(g) + SO2(g) ΔH = -518 KJH2(g) + ½ O2(g) à H2O(g) ΔH = -242 KJ1/8 S8(rhombic) + O2(g) à SO2(g) ΔH = -297 KJ
What is the change in heat (in kJ to one decimal place) when 1.75 L of H2O condenses under 1 atm pressure at 100 °C? ΔHvap = 40.7 kJ mol–1 d(H2O) = 0.958 g mL–1 (at 100 °C)
1) From the following data, C(graphite) + O2(g) à CO2(g) = -393.5 kJ/mol H2(g) + 1/2O2(g) = H2O(l) delta H = -285.8 kJ/mol 2C2H6(g) + 7O2(g) = 4CO2(g) + 6H2O(l) delta H = -3119.6 kJ/mol calculate the enthalpy change for the reaction 2C(graphite) + 3H2(g) =C2H6(g) a) –2417.2 kJ b) –84.6 kJ c) 2417.2 kJ d) –3204.2 kJ
what is the enthalpy of this equation