University:
Miami University
Course:
CHM 141 | College Chemistry
Academic year:

2021
Views:

287

Pages:

3
Author:

Charlee Buchanan

Name: ______________________________________________ Period:______ Due Date:_________ Bond Energy and the Types of Reactions Homework Use the bond enthalpies in the following table to answer the questions which follow: H―H C―H Cl―Cl F―F C―Br 436 414 242 159 285 N≡N O―H C―Cl H―F H―Br 945 463 327 567 366 O=O C=O H―Cl C―F N=O 498 804 431 489 632 N―H C―O 391 358 1. Nitrogen oxide (NO), is formed during electrical storms, The equation for this reaction is: ½ N2(g) + ½ O2(g)  NO(g) Complete each of the tasks to calculate the bond enthalpy for the above reaction. a) Complete the table to show the energy absorbed by each of the bonds in reaction when they are broken Reactant Number of bonds N2 0.5 x N ≡ N O2 0.5 x O=O Enthalpy of bonds broken 0.5 x 945 = kJ mol-1 0.5 x 498 = kJ mol -1 Total enthalpy 472.5 kJ 249 kJ b) Calculate the total energy required 472.5 kJ + 249 kJ = 721.5 kJ c) Complete the table to show the energy released by each of the bonds in reaction when they are formed Product Number of bonds NO 1 x N=O Enthalpy of bonds formed 632 d) How much energy is released? 632 e) Calculate the enthalpy change in the reaction. 721.5 – 632 = 89.5 kJ Total enthalpy 632 Name: ______________________________________________ Period:______ Due Date:_________ Bond Energy and the Types of Reactions 2. Calculate the enthalpy change for each of the following reactions: a) N2(g) + 3H2(g)  2NH3(g) Bonds broken: 3 x H―H = 3 x 436 = 1308 N≡N = 945 Total energy required to break bonds = 1308 + 945 = 2253 kJ Bonds formed: 6 x N―H = 6 x 391 = 2346 Total enthalpy change: 2253 – 2346 = -93 kJ mol-1 b) CH4(g) + Cl2(g) CH3Cl(g) + HCl(g) Bonds broken: 4 x C―H = 4 x 414 = 1656 Cl ―Cl = 242 Total energy required to break bonds = 1656 + 242 = 1898 kJ Bonds formed: 3 x C―H = 3 x 414 = 1242 1 x C―Cl = 327 1 x H―Cl = 431 Total energy released = 1242+ 327 + 431 = 2000 kJ Total enthalpy change = 1898 – 2000= -102 kJ mol-1 c) CH4(g) + F2(g)  CH3F(g) + HF(g) Bonds broken: 4 x C―H = 4 x 414 = 1656 F―F = 159 Total energy required to break bonds = 1656 + 159 = 1815 Bonds formed: 3 x C―H = 3 x 414 = 1242 1 x C―F = 489 1 x H―F = 567 Total energy required to form bonds = Name: ______________________________________________ Period:______ Due Date:_________ Bond Energy and the Types of Reactions 1242 + 489 + 567 = 2298 kJ mol-1 Total enthalpy change = 1815 – 2298 = -483 kJ mol-1 d) CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) Bonds broken: 4 x C―H = 4 x 414 = 1656 2 x O=O = 2 x 498 = 996 Total energy needed to break bonds = 1656 + 996 = 2652 kJ mol-1 Bonds formed: 2 x C=O = 2 x 804 = 1608 4 x O―H = 4 x 463 = 1852 Total energy released when bonds form = 1608 + 1852= 3460 kJ mol-1 Total enthalpy change = 2652 – 3460 = -808 kJ mol-1 3. The reaction between bromine and methane is as follows: CH4(g) + Br2(g)  CH3Br(g) + HBr(g) If the enthalpy change for the reaction is -45 kJ mol-1, calculate the bond energy for Br―Br. Bonds broken: 4 x C―H = 4 x 414 = 1656 EBr―Br = ? Bonds formed: 3 x C―H = 3 x 414 = 1242 C―Br = 285 H―Br = 366 Total bonds formed = 1242 + 285 + 366 = 1893 kJ mol-1 EBr-Br = -45 – 1893 – 1656 EBr-Br = 192 kJ mol-1