g CO 2 g 2 H 2 Ol H Utilizing Hesss law Reactants Standard State Elements H H a

G co 2 g 2 h 2 ol h utilizing hesss law reactants

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(g) CO 2 (g) + 2 H 2 O(l) H = ? Utilizing Hess’s law: Reactants Standard State Elements H = H a + H b = 75 + 0 = 75 kJ Standard State Elements Products H = H c + H d = –394 – 572 = –966 kJ __________________________________________________________________ Reactants Products H = 75 – 966 = –891 kJ b. The standard enthalpy of formation for an element in its standard state is given a value of zero. To assign standard enthalpy of formation values for all other substances, there needs to be a reference point from which all enthalpy changes are determined. This reference point is the elements in their standard state which is defined as the zero point. So when using standard enthalpy values, a reaction is broken up into two steps. The first step is to calculate the enthalpy change necessary to convert the reactants to the elements in their standard state. The second step is to determine the enthalpy change that occurs when the elements in their standard state go to form the products. When these two steps are added together, the reference point (the elements in their standard state) cancels out and we are left with the enthalpy change for the reaction.
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CHAPTER 9 ENERGY, ENTHALPY, AND THERMOCHEMISTRY 379 c. This overall reaction is just the reverse of all the steps in the part a answer. So H = +966 – 75 = 891 kJ. Products are first converted to the elements in their standard state which requires 966 kJ of heat. Next, the elements in the standard states go to form the original reactants [CH 4 (g) + 2 O 2 (g)] which has an enthalpy change of 75 kJ. All of the signs are reversed because the entire process is reversed. 67. In general: Δ H° = o products , f p H Δ n , Δ H n o reactants f, r and all elements in their standard state have o f H Δ = 0 by definition. a. The balanced equation is: 2 NH 3 (g) + 3 O 2 (g) + 2 CH 4 (g) 2 HCN(g) + 6 H 2 O(g) Δ H° = (2 mol HCN × o HCN , f H Δ + 6 mol H 2 O(g) o O H , f 2 H Δ ) (2 mol NH 3 × o NH , f 3 H Δ + 2 mol CH 4 × o CH , f 4 H Δ ) Δ H° = [2(135.1) + 6( 242)] [2( 46) + 2( 75)] = 940. kJ b. Ca 3 (PO 4 ) 2 (s) + 3 H 2 SO 4 (l) 3 CaSO 4 (s) + 2 H 3 PO 4 (l) Δ H° =     mol kJ 1267 ) l ( PO H mol 2 mol kJ 1433 ) s ( CaSO mol 3 4 3 4     mol kJ 814 ) l ( SO H mol 3 mol kJ 4126 ) s ( ) PO ( Ca mol 1 4 2 2 4 3 Δ H° = 6833 kJ (-6568 kJ) = 265 kJ c. NH 3 (g) + HCl(g) NH 4 Cl(s) Δ H° = (1 mol NH 4 Cl × o Cl NH , f 4 H Δ ) (1 mol NH 3 × o NH , f 3 H Δ + 1 mol HCl × o HCl , f H Δ ) Δ =   mol kJ 314 mol 1     mol kJ 92 mol 1 mol kJ 46 mol 1 Δ H° = 314 kJ + 138 kJ = 176 kJ d. The balanced equation is: C 2 H 5 OH(l) + 3 O 2 (g) 2 CO 2 (g) + 3 H 2 O(g) Δ =     mol kJ 242 mol 3 mol kJ 5 . 393 mol 2   mol kJ 278 mol 1 Δ = 1513 kJ ( 278 kJ) = 1235 kJ
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380 CHAPTER 9
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