Question

For which of the following reactions is AHºrx equal to the heat of formation for the product? a) N2(g) + 3H2(g) + 2NH3(g) b) Na (s) + 2Cl2 (1) NaCl (s) c) CH. (1) - 6C (s, graphite) + 3H2(g) X d) VP4 (s, white) + 2H2(g) + Br2 (1) - PH Br (1) e) 12C (g) + 11H2(g) + 110 (g) → CH22011 (g) X 6) Ammonia burns in air as follows: 4NH3(g) + 302 (g) 2N2 (g) + 6H20 (g). How many liters of N2 (9 formed by burning 2.50 L of ammonia in excess oxygen with all species at the same temperature and pressure? a) 2.00 L b) 2.50 L c) 1.25 L d) 5.00 L e) none of these 288

Answer 1

Standard enthalpy of formation is the change of enthalpyduring the formation of 1 mole of substance from its constituent elements, with all substancesin their standard states. (a) N2 (g) 3 H2 (g) ----> 2 NH3 (g) Since 2 moles of NH3 are formed, so AH°rxn is not equal to heat of formation of product. (b) Na (s) Cl2 (I) Cl2 (I) is not standard state of chlorine, so AH°rxn is not equal to heat of formation of product NaCl (s) (c) This is not formation reaction (d) P (s, white) 2 H2 (g) Br2 () > PH4Br (I) In this 1 moles of PH4Br is forming from its constituent elements. So, AH°rxn is equal to heat of formation of product (e) O (g) is not standard state of chlorine, so AH°rxn is not equal to heat of formation of product Option D is the right answer.

The balanced equation is 4 NH3 (g)3 02 (g) ----- --> 2 N2 (g) 6 H20 (g) We know that PV nRT At constant T and P, V oc n Accordingto balanced equation, 4 moles of NH3 produces 2 moles of N2 That is 4 L of NH3 produces 2 L of N2 So, 2.50 L of NH3 would produce 2.50 L 2/4 1.25 L of N2 Volume of N2= 1.25 L Option C is the right answer.