Hess’s Law SECTION 5.3. Hess’s Law  The enthalpy change of a physical or chemical process depends only on the initial and final conditions of the process.

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Presentation transcript:

Hess’s Law SECTION 5.3

Hess’s Law  The enthalpy change of a physical or chemical process depends only on the initial and final conditions of the process.  The enthalpy change of a multistep process is the sum of the enthalpy changes of its individual steps.

Problem  Iron can be obtained from the following reaction:  Fe 2 O 3(s) + 3CO (g) → 3CO 2 + 2Fe (s)  Determine the enthalpy change of reaction, given the following: 1. CO (g) + ½ O 2 → CO 2 ∆H°=-283.0kJ 2. 2Fe (s) + 3/2 O 2 → Fe 2 O 3(s) ∆H°= kJ

Solution  Manipulate equations to make them match the overall equation:  #1 equation proceeds in the correct direction but must be multiplied by 3.  So:  CO (g) + ½ O 2 → CO 2 ∆H°=-283.0kJ X 3  3CO (g) + 3/2 O 2 → 3CO 2 ∆H°= kJ  #2 equation needs to be reversed but the coefficients are ok.  So:  2Fe (s) + 3/2 O 2 → Fe 2 O 3(s) ∆H°= kJ becomes:  Fe 2 O 3(s) → 2Fe (s) + 3/2 O 2 ∆H°= 824.3kJ (note the sign is reversed)

 To get the overall equation:  Since oxygen is the same on both sides of the equation, cross it out and add the individual enthalpy changes:  Fe 2 O 3(s) + 3CO (g) → 3CO 2 + 2Fe (s) ∆H°= kJ

Practice Problems – p. 316 – all questions

Standard Molar Enthalpies of Formation  Change in enthalpy when 1 mole of a compound is formed directly from its elements in their most stable state at SATP (25°C & 100kPa).  It’s a synthesis reaction when the compound is formed directly from its elements and not from any other compound.  Eg. C (s) + O 2(g) → CO 2 (g)  Coefficients are often fractions because there must be only 1 mole of product formed.  See Table 5.5 and Appendix B

Thermal Stability  The ability of a substance to resist decomposition when heated.  The greater the enthalpy change of a decomposition reaction, the greater its thermal stability.

Enthalpies of Formation and Hess’s Law  ∆H° r =∑(n∆H° f products) - ∑(n ∆H° f reactants)  n= coefficient in equation  ∑ = “sum of”  ∆H° r = enthalpy change of reaction  Use standard molar enthalpies of formation on Appendix B

Determine the enthalpy of formation for the following reaction :  CH 4(g) + 2O 2 → CO 2 + 2H 2 O  ∆H° r =∑(n∆H° f products) - ∑(n ∆H° f reactants) **Use Appendix B  ∆H° r =((1)(-393.2) + (2)(-241.9)) - ((1)(-74.6)(2)(0)) ** since oxygen is an element in its most stable state, standard molar enthalpy is 0.  = (-877.1kJ)-(-74.6kJ)  = kJ

Videos/Resources to study at home  Hess’s Law:    Hess’s Law and Enthalpies of Formation  

Practice Problems – p. 323 – #51, 52, 54, 55, 56, 58.  Research Application Questions  Choose from:  P. 291: #4, 10, 12  P. 311: #8, 12  P. 335: #2, 5, 6, 7, 8  P. 349: #62, 63, 65, 66, 67, 69, 70  P. 351: #23  P : #68, 69