1)Given the following reaction: H2(g)+I2(s) ? 2HI(g) with a ? H of 52.9 kJ. What
ID: 808723 • Letter: 1
Question
1)Given the following reaction: H2(g)+I2(s) ? 2HI(g) with a ? H of 52.9 kJ. What is the change in enthalpy for the following reaction: HI(g) ? H2(g)+ I2(s)? Express your answer in kJ.
2)How much heat is released during the formation of 3.78 mol HCl(g) in this reaction: H2(g)+Cl2(g) ? 2HCl(g) with a ? H of -184.6 kJ. Express your answer in kJ.
3)What will be the final temperature if a 4.00 g silver ring at 41.0 C if it gives off 22.0 J of heat to the surroundings. The specific heat capacity of silver is 0.235 J/(g C). Express your answer in C.
4)Let's say you hold some ice in your hand. The ice melts and your hand feels cold. Is the reaction H2O(s) ? H2O(l) endothermic or exothermic? Is the change in enthalpy of the reaction positive or negative.
exothermic, positive
endothermic, negative
endothermic, positive
exothermic, negative
5)B5H9(l) is a colorless liquid that will explode when exposed to oxygen. How much heat is released when 0.211 mol of B5H9 reacts with excess oxygen where the products are B2O3(s) and H2O(l). The standard enthalpy of formation of B5H9(l) is 73.2 kJ/mol, the standard enthalpy of formation of B2O3(s) is -1272 kJ/mol and that of H2O(l) is -285.4 kJ/mol. Express your answer in kJ. Enter only a numerical value, do not enter units.
6)When 0.105~mol propane, C3H8 is burned in an excess of oxygen, how many moles of oxygen are consumed? The products are carbon dioxide and water. You have to properly balance the equation before you can answer this question.
0.420~mol O2
0.905~mol O2
0.875~mol O2
0.525~mol O2
0.975~mol O2
Explanation / Answer
(1) The given thermochemical reaction is
H2(g)+I2(s) ------> 2HI(g) DeltaH1 = +52.9 kJ -------(1)
Now we need to find DeltaH(rxn) for the following reaction
HI(g) --------> 1/2 H2(g) + 1/2 I2(s) DeltaH(rxn) = ?---------(2)
Since we need one molecule of HI(g) on reactant side in rxn(2), we can achieve this by reversing rxn(1) and then dividing by 2. When we reverse the chemical equation, DeltaH will become negative and when we divide the chemical reaction by 2, the DeltaH value also needs to be divided by 2.
Hence DeltaH(rxn) for the above reaction can be calculated as
DeltaH(rxn) = -(+52.9 kJ)x1/2 = - 26.45 KJ (answer)
(2) The given thermochemical reaction is
H2(g)+Cl2(g) ----à 2HCl(g), DeltaH = -184.6 kJ
Since the reaction is exothermic, 184.6 kJ of heat is released during the formation of 1 mol of HCl.
Hence the amount of heat released during the formation of 3.78 mol of HCl(g)
= (184.6kJ)x(3.78/1) = 698 kJ (answer)
(3) Given mass of silver ring, m = 4.00 g
Initial temperature, Ti = 41 DegC
Let the final temperature be Tf
Heat given off to the surrounding, Q = - 22.00 J
Specific heat capacity of silver, s = 0.235Jg-1C-1
When heat is given off to the surrounding the temperature will decrease.
Now applying the formulae
Q = m*s*DeltaT = m*s*(Tf – Ti)
=> Tf – Ti = Q / m*s = -22.00J / 4.00g*( 0.235Jg-1C-1)
=> Tf – 41C = -23.40C
=> Tf = - 23.40 C+ 41C = 17.6 degC (answer)
(4) Since our hand feels cold, heat is reJeased by our hand and this heat is absorbed byice. When this absorbed heat overcomes the latent heat of fusion of ice, ice starts melting. Hence the process of melting of ice is endothermic process and the DeltaH for the reaction is +ve.
Hence third option is correct option.
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