Given : (a) BC13(2) + 3H2O¶→H;BO3(s) + 3HClg) AH = -112.5 kJ (b) B2H6(g) + 6H2O→2H3BO3(s) + 6H2(g) AH = -493.4 kJ (c) 1/2H2(g) + 1/2C12(g) →HClg) AH = -92.3 kJ Calculate the value of AH for the reaction: B2H6(g) + 6Cl2g) →2BC13(g) + 6HClg) AH = ? Select one: a. -1376KJ O b. -1476KJ O c. -1576KJ O d. -1576KJ

Given : (a) BC13(2) + 3H2O¶→H;BO3(s) + 3HClg) AH = -112.5 kJ (b) B2H6(g) + 6H2O→2H3BO3(s) + 6H2(g) AH = -493.4 kJ (c) 1/2H2(g) + 1/2C12(g) →HClg) AH = -92.3 kJ Calculate the value of AH for the reaction: B2H6(g) + 6Cl2g) →2BC13(g) + 6HClg) AH = ? Select one: a. -1376KJ O b. -1476KJ O c. -1576KJ O d. -1576KJ

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...

See similar textbooks

Similar questions

N2(g) + O2(g)< -----> 2 NO(g) H = +180.7 kJ 2 H2O(l) <-----> 2 H2(g) + O2(g) H = +571.7 kJ 4 NH3(g) + 5 O2(g) <-----> 4 NO(g) + 6 H2O(l) H = -1167.1 kJ Use Hess's law to calculate H for the reaction below. 2 NH3(g) <-----> N2(g) + 3 H2(g)
Using the following thermochemical equations; (a) CaO (s) + Cl2 (g) → CaOCl2 (s) ΔH°= -110.9 kJ(b) CaO (s) + H2O (l) → Ca(OH)2 (s) ΔH°= -65.1 kJ(c) H2O (l) + CaOCl2 (s) + Br2 (l) → Ca(OH)2 (s) + Br2 (l) +2 NaCl (s) ΔH°= -60.2 kJDetermine the value of ΔH° in kJ for the following reaction:½ Cl2 (g) + NaBr (s) → ½ Br2 (l) + NaCl (s)
Iron can be extracted from the iron(III) oxide found in iron ores (such as haematite) via an oxidation-reduction reaction with carbon. The thermochemical equation for this process is: 2 Fe203(s) + 3 C(s) → 4 Fe(1) + 3 CO2(g) AH° = +467.9 kJ
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)
A chemist measures the enthalpy change AH during the following reaction: 4 Fe(s) + 3 02(g)→2 Fe,O3(s) AH =-320. kJ Use this information to complete the table below. Round each of your answers to the nearest kJ/mol. reaction ΔΗ x10 Fe(s) + 0,6) - Fe,0,(1) O kJ 2Fe,0; (s) – 4Fe(s) + 30, (3) O kJ 4Fe,0, (s) - 8Fe(s) + 60, (3)
Given the reactions, X(s)+12O2(g)⟶XO(s) Δ?=−909.7 kJX(s)+12O2(g)⟶XO(s) ΔH=−909.7 kJ XCO3(s)⟶XO(s)+CO2(g)Δ?=+350.1 kJXCO3(s)⟶XO(s)+CO2(g)ΔH=+350.1 kJ what is Δ?ΔH for this reaction? X(s)+12O2(g)+CO2(g)⟶XCO3(s)X(s)+12O2(g)+CO2(g)⟶XCO3(s) Δ?= kJ
The enthalpy changes, AH, for three reactions are given. H, (g) + H, O(1) AH= -286 kJ/mol Ca(s) + 2 H*(aq) Ca2+(aq) + H, (g) Ca+(aq) + H, O(1) AH = -544 kJ/mol CaO(s) + 2 H*(aq) AH = -193 kJ/mol Using Hess's law, calculate the heat of formation for CaO(s) using the reaction shown. Ca(s) + O,(g) CaO(s) AH kJ/mol
A chemist measures the enthalpy change AH during the following reaction: 2 KCl(s) + 302(g) →2 KClO3(s) AH=78. kJ Use this information to complete the table below. Round each of your answers to the nearest kJ. reaction ΔΗ ㅁ x10 2K C103 (s) → 2KC1(s) + 302(g) KCIO, (s)-KCI(s) + 0,(s) 3 KCI(s)+0, (g)KCIO, (s) kJ kJ 1 5
27. When 0.0500 mol of CH4 (g) burns, 40.1 kJ of heat is produced. What is the change in enthalpy (AH) in kJ for the following reaction: CH4 (g) (a) +802 kJ/mol + 202 (g) (b) -802 kJ/mol , CO₂ (g) + 2H₂O(g) + Heat (c) +40.1 kJ/mol (d) -40.1 kJ/mol
Given the following thermochemical reactions: N2H4() + 2H2O2(g) → N2(g) + 4H20() A) AH° = -818.3 kJ В) H2(g) + ) ½O2(g) → H20() AH° = -285.8 kJ C) H2(g) + 02(g) → H202(g) AH° = -187.8 kJ Calculate AH° of the reaction: N2H4() + O2(g) → N2(g) + 2H20() Do not encode the units. Input the final answers with 2 decimal places.
Calculate AHO,xn for each of the following using data from Appendix B. (a) 2 H2S(g) + 3 02(g) → 2 SO2(g) + 2 H20(g) kJ (b) SiO2(s) + 4 HF(g) → SİF4(g) + 2 H20(/) kJ
Find the enthalpy change for this reaction using the data below. Clearly show how you got the answer. 2 Al(s) + 3 Cl2 (g) → 2 AlCl3 (s) Data: (a) 2 Al(s) + 6 HCl(aq) → 2 AlCl3 (aq) + 3 H2 (g) H = −1049 kJ (b) AlCl3 (s) → AlCl3 (aq) H = −323 kJ (c) H2 (g) + Cl2 (g) → 2 HCl(g) H = −185 kJ (d) HCl(aq) → HCl(g) H = +74.8 kJ