Chapter 15 - Chemical Thermodynamics - Exercises - Thermochemical Equations and Hess's Law - Page 600: 32

Equation (1) $4HCl_{(g)} + O_{2(g)} →2H_{2}O{(l)}+2Cl_{2(g)}$ $ \Delta H(1) = - 202.4kJ/mol$ Equation (2) $\frac{1}{2}H_{2(g)} + \frac{1}{2} F_{2(g)} →HF{(l)}$ $ \Delta H(2) = - 600.0kJ /mol$ Equation (3) $H_{2(g)} + \frac{1}{2} O_{2(g)} →H_{2}O{(l)}$ $ \Delta H(3) = - 285.8kJ /mol$ Divide Equation 1 by 2 to get Equation (4) Equation (4) $2HCl_{(g)} + \frac{1}{2} O_{2(g)} →H_{2}O{(l)}+Cl_{2(g)}$ $ \Delta H(4) = - 101.2kJ/mol$ Multiply Equation 2 by 2 to get Equation (5) Equation (5) $H_{2(g)} + F_{2(g)} →2HF{(l)}$ $ \Delta H(5) = - 1200.0kJ /mol$ Reverse Equation 3 to get Equation 6 Equation 6 $H_{2}O{(l)} → H_{2(g)} + \frac{1}{2} O_{2(g)} $ $ \Delta H(6) = +285.8kJ /mol$ The required equation, $2HCl_{(g)} + F_{2(g)} →2HF{(l)}+Cl_{2(g)}$is formed by adding Equation (4), Equation (5) and Equation (6). $ \Delta Hrxn = \Delta H(4) +\Delta H(5) +\Delta H(6)= (-101.2 + - 1200.0 + 285.8 ) kJ/mol = - 1015.4 kJ/mol $

Equation (1) $4HCl_{(g)} + O_{2(g)} →2H_{2}O{(l)}+2Cl_{2(g)}$ $ \Delta H(1) = - 202.4kJ/mol$ Equation (2) $\frac{1}{2}H_{2(g)} + \frac{1}{2} F_{2(g)} →HF{(l)}$ $ \Delta H(2) = - 600.0kJ /mol$ Equation (3) $H_{2(g)} + \frac{1}{2} O_{2(g)} →H_{2}O{(l)}$ $ \Delta H(3) = - 285.8kJ /mol$ Divide Equation 1 by 2 to get Equation (4) Equation (4) $2HCl_{(g)} + \frac{1}{2} O_{2(g)} →H_{2}O{(l)}+Cl_{2(g)}$ $ \Delta H(4) = - 101.2kJ/mol$ Multiply Equation 2 by 2 to get Equation (5) Equation (5) $H_{2(g)} + F_{2(g)} →2HF{(l)}$ $ \Delta H(5) = - 1200.0kJ /mol$ Reverse Equation 3 to get Equation 6 Equation 6 $H_{2}O{(l)} → H_{2(g)} + \frac{1}{2} O_{2(g)} $ $ \Delta H(6) = +285.8kJ /mol$ The required equation, $2HCl_{(g)} + F_{2(g)} →2HF{(l)}+Cl_{2(g)}$is formed by adding Equation (4), Equation (5) and Equation (6). $ \Delta Hrxn = \Delta H(4) +\Delta H(5) +\Delta H(6)= (-101.2 + - 1200.0 + 285.8 ) kJ/mol = - 1015.4 kJ/mol $