Question 1
When benzene vapor condenses at constant pressure, the sign of the enthalpy change for the process
is negative.
Energy is required for vapourization (positive enthalpy change) and therefore energy is given off upon condensation (negative enthalpy change).
Question 2
The enthalpy of sublimation of mercury is 61.6 kJ/mol and its enthalpy of vaporization is 59.3 kJ/mol. Calculate the enthalpy of fusion.
+2.3 kJ/mol
The enthalpy of fusion is the enthalpy for melting one mole of a substance and is always positive. Because enthalpy is a state function, the enthalpy of sublimation is equal to the enthalpy of fusion plus the enthalpy of vapourization.
Question 3
Calculate the reaction enthalpy for the decomposition of hydrogen peroxide,
2H2O2(l) --> 2H2O(l) + O2(g)from the data
H2(g) + O2(g) --> H2O2(l) DHo = -187.8 kJ .....(1)
2H2(g) + O2(g)--> 2H2O(l) DHo = -571.6 kJ ....(2)
-196.0 kJ
Reverse equation (1) and multiply it by 2 (remember to change the sign of DHo and multiply it by 2); add to equation (2).
DHo = 2 x (+187.8) + (-571.6) = -196.6 kJ/mol
Question 4
If the standard enthalpy of combustion of glucose, C6H12 O6(s), is -2808 kJ/mol, calculate the heat output if 250 g of glucose is burned.
3.90 x 103 kJ
mol glucose = 250 g/180 g/mol
heat output = (250/180)mol x 2808 kJ/mol = 3900 kJ
Question 5
** There is a programming problem with this question. Sorry for the inconvenience, however the correct answer is shown below. **
Which thermochemical equation gives the value of the standard enthalpy of formation for HI(g)?1/2H2(g) + 1/2I2(s) --> HI(g)
The standard enthalpy of formation of a substance is the enthalpy change for the formation of 1 mole of substance from its elements in their most stable forms at 1 atm and a specified temperature, usually 25oC.
Question 6
Which of the following has a standard enthalpy of formation equal to zero?
Hg(l)
The standard enthalpy of formation of an element in its most stable form is zero. Mercury is a liquid at 1 atm and 25o. The most stable forms of sodium, carbon, iodine and sulfur are the solids.
Question 7
When SnO2(s) is formed from the combustion of gray tin, the reaction enthalpy is
-578.6 kJ, and when white tin is burned to form SnO2(s), the reaction enthalpy is
-580.7 kJ. Calculate the reaction enthalpy for
Sn(gray) -->Sn(white)
+2.1 kJ
The required equation is obtained by adding the equation for the combustion of Sn(gray) to the reverse of the equation for the combustion of Sn(white).
DHrxn = -578.6 + (+580.7) = +2.1 kJ/mol
Question 8
The standard enthalpies of combustion of diamond and graphite
are -395.41 and -393.51 kJ/mol, respectively.
What is the standard enthalpy of formation of diamond?
+1.90 kJ/mol
The equation for the standard enthalpy of formation of diamond is
C(graphite) --> C(diamond).
The latter can be obtained from the combustion equations.
DHfo = -393.51 + (+395.41) = +1.90 kJ/mol.
Question 9
Calculate DU for a system that absorbs 350 kJ of heat and has 625 kJ of work done on the system.
+975 kJ
Heat supplied to the system is positive and work done on the system is positive. Therefore, DU = q + w = +350 + (+625) = +975 kJ.
Question 10
Calculate the work needed to make room for products in the combustion of 1 mole of CH4(g) to carbon dioxide and water vapor at STP. (1 L atm = 101 J)
no work is needed
The equation for the combustion of one mole of CH4(g) shows that the number of moles of gaseous reactants equals the number of moles of gaseous products. Therefore, DV = 0 and no work is done.