Part A) Calculate the enthalpy change, H, for the process in which 25.4 g of wat
ID: 530045 • Letter: P
Question
Part A) Calculate the enthalpy change, H, for the process in which 25.4 g of water is converted from liquid at 16.5 C to vapor at 25.0 C . For water, Hvap = 44.0 kJ/mol at 25.0 C and Cs = 4.18 J/(gC) for H2O(l). Express your answer to three significant figures and include the appropriate units.
Part B) How many grams of ice at -30.0 C can be completely converted to liquid at 23.4 C if the available heat for this process is 5.38×103 kJ ? For ice, use a specific heat of 2.01 J/(gC) and Hfus=6.01kJ/mol. Express your answer to three significant figures and include the appropriate units.
Part C)
Calculate the standard enthalpy change for the reaction
2A+B2C+2D
Use the following data:
Part D)
Consider the exothermic reaction
2C2H6(g)+7O2(g)4CO2(g)+6H2O(g)
Calculate the standard heat of reaction, or Hrxn, for this reaction using the given data. Also consider that the standard enthalpy of the formation of elements in their pure form is considered to be zero.
Substance Hf(kJ/mol) A -245 B -391 C 209 D -515
Explanation / Answer
1.
Total heat to be supplied= sensible heat + latent heat= 0.903+61.6= 62.503 KJ
2.
Let the mass of ice be m
Total heat to be supplied= 0.0603m+0.333m +0.09791m= 5.38*103
Hence 0.491m= 5.38*1000, m= 10957 gm of water= 10.957 kg of water.
3.
Standard enthalpy change= sum of standard enthalpy of products- sum of standard enthalpy of reactants
=2* standard enthalpy of C+2* standard enthalpy of D- { 2* standard enthalpy of A + 1* standard enthalpy of B)
2,2,2 and 1 are coefficients of C,D, A and B in the reaction.
Hence standard enthalpy change= 2*209+2*(-515)- { 2*(-245)-391}=269 Kj
4.
Standard enthalpy change= sum of standard enthalpy of products- sum of standard enthalpy of reactants . For the reaction
2C2H6(g)+7O2(g)4CO2(g)+6H2O(g)
=4* standard enthalpy of formation of CO2+6* standard enthalpy of formation of H2O- { 2* standard enthalpy of formation of C2H6 + 7* standard enthalpy of O2).
Standard enthalpy change of O2=0
=4*(-393.5)+6*(-241.8)- {2*(-84.7)+7*0} = -2855.4 KJ