From the enthalpies of reactio:
H2(
From the enthalpies of reactio: H2( 9.) From the enthalpies of reaction Hale) + F(b) →...
Calculate AH for the reaction CyHa(s) + 6F2(8) 2CF4(8) + 4HF(G) given the following information: As part of your answer, show how the equations can be added together to give the overall or net equation H2(g) + F2(g) + 2HF(g) AH--537 kJ C(s) + 2F2(8) CF4(8) AH--680. kJ 2C(s) + 2H2(g) → C2H4(8) AH- +52.3kJ HTML Editore BIVA-A- IEE311xx, - E V O 1 12pt - Paragraph -
can you explain why the answer is correct? QUESTION 11 Given the information below calculate AHrxn for: C2Halg)+6F2g)> 2CF4 + 4HF(g) => 2HF(g ) AH°= -537 kJ ->CF4(g ) A °H-680 kJ H2g)+F2(g) C(s)+2F29) 2C(s)+2H2(g)=> C2H4(g) AH 52.3 kJ A 1165 kJ B..1269 C.-2382 kJ D.-2486 kJ E. 1642 kJ
5.65 From the enthalpies of reaction H 2 (g)+r 2 (g) longrightarrow2 HF(s) Delta*H = - 537kJ C(s)+2 F 2 (s) longrightarrow CF 4 (g) Delta*H = - 680kJ 2C(s)+2 H 2 (g) C 2 H 4 (g) Delta*H = + 52.3kJ calculate Delta*H for the reaction of ethylene with F 2 : C 2 H 4 (g)+6 F 2 (g) 2 CF 4 (g)+4 HF(g) 5.65 From the enthalpies of reaction H2(8) + (8) 2 HF(8) AH = -537...
6.65 From the enthalpies of reaction H2(8) + F2(8) 2 HF(8) AH = -537 kJ C(s) + 2 F2(8) CF4(8) AH = -680 kJ 2 C(s) + 2 H2(8) —— C2H4(8) AH = +52.3 kJ calculate AH for the reaction of ethylene with F2: C2H4(8) + 6 F2(8) — 2 CF4(8) + 4 HF(8)
18. Given that AHorxn = -546.6 kJ H2 (g)+F2 (g)2HF (g) AH°rxn = -571.6 kJ 2H2 (g)+ O2 (g) 2H20 (1) Calculate the value of AHrxn for 2F2 (g) 4HF (g) + O2 (g) +2H20 (I)
2. Calculate AH for the reaction of ethylene with F2. CHA(g) + 6 F2(g) → 2 CF.(g) + 4 HF(g) using the following enthalpies of reaction: 2 H2(g) + 2 F2(g) → 4 HF(g) AH, = -1074 kJ C(s) + 2 F2(g) → CF4(g) AH2 = -680 kJ C(s) + H2(g) → CH.(g) AH, = 26.2 kJ
A chemist knows that the kJ for the reaction 2H2(g) + O2 (g) ---> 2H2O (g) ,and that kJ for the reaction H2 (g) + F2 (g) ---> 2HF (g). With this information he calculated the for the reaction 2H2O (g) + 2F2 (g) ---> 4HF(g) + O2 (g) and predicted whether was positive or negative. How? A Ho- 485
2. Use Hess’s Law to determine the enthalpy of the reaction below. 2F2(g) + 2H2O(l) → 4HF(aq) + O2(g) DH˚= ? H2(g) + F2(g) → 2HF(aq) DH˚ = -546.6 kJ 2H2 (g) + O2(g) → 2H2O(l) DH˚ = -571.6 kJ a. 42 kJ b. -1120 kJ c. -251 kJ d. -521 kJ e. -1690 kJ
A scientist measures the standard enthalpy change for the following reaction to be - 87.7 kJ : 2HBr(g) + Cl2(g)— 2HCl(g) + Bra(s) Based on this value and the standard enthalpies of formation for the other substances, the standard enthalpy of formation of HCl(g) is kJ/mol CH4() + H2O(g)_3H2(g) + CO(g) Based on this value and the standard enthalpies of formation for the other substances, the standard enthalpy of formation of H2O() is kJ/mol Given the standard enthalpy changes for...
2) From the enthalpies of reaction provided, calculate the AH for the reaction of carbon with hydrogen as shown below. Is the reaction exothermic or endothermic? How many grams of C3Hg are produced with an enthalpy change of 209.4 kJ? 3 C(s) + 4 H2(g) - C He(a) moto-non (5+3=2 C3H8(9) + 5 O2(g) 3 CO2(g) + 4 H2O(9) AH = -2043 kJ/mol AH = -393.5 kJ/mol C(s) + Ozon 2 H2(g) + O2th CO2(g) 2 H20(9) AH = -483.6...