Chp3_DH_f-1

Chp3_DH_f-1 - CaO CaCO → 29 kJ CaO H f 1 635-= ∆ 29 kJ...

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Calculating H r from “Heats of Formation” For any chemical reaction, calculation of the enthalpy change is often possible from tabulated “heats of formation:” namely, enthalpies of (formation) reactions. Equation 3.11 is the operative expression. Examples 3.10 and 3.11 apply to this and one must use this approach for WA Q06 problems 3-6 (3.11 is especially pertinent to problem 5.) Note that the heat of formation for any element in its standard state is zero. Example: Calculate Hº for the reaction: 2 3 CO
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Unformatted text preview: CaO CaCO + → ( 29 kJ CaO H f 1 . 635-= ∆ ( 29 kJ CaCO H f 9 . 1206 3-= ∆ ( 29 kJ CO H f 5 . 393 2-= ∆ Using equation 3.11, we obtain: ( 29 kJ H 3 . 178 9 . 1206 1 . 635 5 . 393 + =----= ∆ . Note, that this is the enthalpy change for the reaction AS WRITTEN . Now, suppose that 4.32g of CaCO3 reacts. What is ∆ H for this actual reaction ? 3 3 3 0432 . 1 . 101 1 32 . 4 CaCO mol g CaCO mol CaCO g = × and thus: kJ CaCO mol kJ CaCO mol 69 . 7 1 3 . 178 0432 . 3 3 = + ×...
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This note was uploaded on 03/30/2008 for the course CH 201 taught by Professor Warren during the Fall '07 term at N.C. State.

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