PS3_Key - Chemistry 2080 Spring 2009 Problem Set #3 - due...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Chemistry 2080 Spring 2009 Problem Set #3 - due date: Friday February 13th at 2 pm Name:____( KEY)_______ Lab TA Name:__________________ Lab Day/Time:______________________ Practice problems (not graded). Chapter 7: 39, 45, 56, 59, 71, 97. 1) Given the following thermochemical data (all at 25°C): (1) . . . . . . .NH 4 Cl(s) + HCl(g) –––––––> H 2 (g) + Cl 2 (g) + NH 3 (l) Δ rxn = +336.5 kJ (2) . . . . . . .NH 3 (g) + HCl(g) ––––––––> NH 4 Cl(s) Δ rxn = – 176.2 kJ The standard enthalpy of vaporization of NH 3 is + 23.4 kJ/mol. The standard enthalpy of formation of NH 3 (g) is – 46.1 kJ/mol Calculate Δ rxn for the reaction: N 2 (g) + 4H 2 (g) + Cl 2 (g) –––––––> 2NH 4 Cl(s) ******************************************* Let’s call the two chemical equations (above) equations (1) and (2). Also, writing out the equations for the standard enthalpy of vaporization and the standard enthalpy of formation of ammonia (and calling them equations (3) and (4), we get: (3) . . . . NH 3 (l) –––– > NH 3 (g) Δ vap = +23.4 kJ (4) . . . . N 2 + 3H 2 –––––> 2NH 3 (g) Δ f = 2x(– 46.1 kJ/mol) = –92.2 kJ If we now add the equations in the following way: (2) + (–1) + (4) + (– 3) we get: NH 3 (g) + HCl(g) +H 2 (g) + Cl 2 (g) + NH 3 (l) + N 2 +3H 2 +NH 3 (g) –––––––> NH 4 Cl(s) + NH 4 Cl(s) + HCl(g) + 2NH 3 (g) + NH 3 (l) which reduces to the equation that we want, namely: N 2 (g) + 4H 2 (g) + Cl 2 (g) –––––––> 2NH 4 Cl(s) So we need to add the standard enthalpies in the same way [i.e. (2) + (–1) + (4) + (– 3)]: i.e. (– 176.2 kJ) + (– 336.5 kJ) + (– 92.2 kJ) + (– 23.4 kJ) = – 628.3 kJ which is Δ rxn for the reaction as written
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
2) As we saw in lecture, the complete combustion of benzene(l) (C 6 H 6 ) in oxygen in a bomb calorimeter resulted in the liberation of 3,268 kJ/mol of heat. The sole products were H 2 O(l) and CO 2 (g). Thus, for the combustion of benzene at 25°C, q v = – 3,268 kJ/mol of benzene. Write a balanced equation for the complete combustion of
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 03/05/2009 for the course CHEM 2080 taught by Professor Davis,f during the Spring '07 term at Cornell University (Engineering School).

Page1 / 4

PS3_Key - Chemistry 2080 Spring 2009 Problem Set #3 - due...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online