DH degree 268.2 kJ for the reaction: FeBr (s) Fe (s) + 3/2 Hr Br_2(I). For this
ID: 1052104 • Letter: D
Question
DH degree 268.2 kJ for the reaction: FeBr (s) Fe (s) + 3/2 Hr Br_2(I). For this reaction, which have, on the average, stronger bonds, the reactants or the products? 1.30 g of a solid were added to a coffee cup calorimeter containing 50.4 g of water at 24.1 degree C. When the solid had dissolved in the water, the temperature of the solution (water) had increased to 30.4 degree C. Find DH for the formation of this solution in Joules per gram of solid dissolved. (The specific heat of water is 4.18 J degree C per J/g degree C.) A reaction between HCI and NaOH was earned out in a coffee cup calorimeter containing 50.0 g of water at 23.5 degree C. DH degree for the reaction is -1442 J. Calculate the temperature of the water when the reaction is over.Explanation / Answer
1]
delta Hrxn = Bond energies of reactants - Bond energies of products
since, delta Hrxn = 268.2 KJ
Reactants have stronger bond energies
2]
q = m*C*dT
q = [50.4+1.3]*4.18*[30.4 - 24.1] = 1361.46J
delta H of solution per gram of solid dissolved = -1361.46 / 1.3 = -1047.28 J
3]
q = m*C*dT
1442 = 50*4.18* [ Tf-23.5]
Tf = 30.399 C
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