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chemsheets answer booklet to chemsheets thermodynamics summary/ revision booklet

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© www.CHEMSHEETS.co.uk 23-Feb-2016 Chemsheets A2 1014 Page 1

, Most answers in this topic are to the nearest unit unless stated otherwise (as
data is to nearest unit and it is addition or subtraction


PAGE 3 EXAMPLES
C(s) + 2 H2(g)  CH4(g) CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(l)
H2(g) + ½ O2(g)  H2O(l) H2(g) + ½ O2(g)  H2O(l)
½ N2(g) + 3/2 H2(g)  NH3(g) C2H6(g) + 3½ O2(g)  2 CO2(g) + 3 H2O(l)
2 C(s) + 3 H2(g) + ½ O2(g)  C2H5OH(l) C2H5OH(g) + 3 O2(g)  2 CO2(g) + 3 H2O(l)
C(s) + 3/2 H2(g) + ½ Br2(l)  CH3Br(l) Na(s) + ¼ O2(g)  ½ Na2O(s)
2 Na(s) + ½ O2(g)  Na2O(s) C6H14(g) + 9½ O2(g)  6 CO2(g) + 7 H2O(l)


HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
½H2SO4(aq) + NaOH(aq)  ½ Na2SO4(aq) + H2O(l)
HNO3(aq) + KOH(aq)  KNO3(aq) + H2O(l)
HNO3(aq) + ½Ba(OH)2(aq)  ½ Ba(NO3)2(aq) + H2O(l)
½H2SO4(aq) + ½Ba(OH)2(aq)  ½BaSO4(aq) + H2O(l)


PAGE 4 EXAMPLES
Ex1  H = [Sum of Hf products] – [Sum Hf reactants]
= [ -394 + 2(-286) ] – [–75 + 0]
-1
= –891 kJ mol


Ex2  H = [Sum of Hf products] – [Sum Hf reactants]
-2877 = [ 4(-394) + 5(-286) ] – [fH + 0]
fH = 2877 + 4(-394) + 5(-286)
-1
= –129 kJ mol


PAGE 5 EXAMPLES

Ex 1 DH CH4(g)
C(s) + 2 H2(g)

–394 –890
2(–286)
CO2(g) + 2 H2O(l)
 H – 890 = –394 + 2(–286)
H = –394 + 2(–286) + 890
-1
= –76 kJ mol




© www.CHEMSHEETS.co.uk 23-Feb-2016 Chemsheets A2 1014 Page 2

, Ex 2 –75 CH3CH2OH(l)
2 C(s) + 3 H2(g) + ½ O2(g)

2(–394) DcH
3(–286)
2 CO2(g) + 3 H2O(l)
 cH – 75 = 2(–394) + 3(–286)
cH = 2(–394) + 3(–286) + 75
-1
= –1571 kJ mol




PAGE 6/7 EXAMPLES

Ex 1 –583 N2(g) + 2 H2O(g)
N2H4(g) + O2(g)
N-N NºN
4(N-H) 4(O-H)
O=O

3 CO2(g) + 4 H2O(l)
 –583 + (NN) + 4(O-H) = (N-N) + 4(N-H) + (O=O)
 (N-N) = –583 + (NN) + 4(O-H) – 4(N-H) – (O=O)
 (N-N) = –583 + 944 + 4(463) – 4(388) – (498)
-1
= +163 kJ mol


Ex 2 -1015 2 CO2(g) + 3 H2O(g)
C2H5OH(l) + 3 O2(g)
44 C-O 4(C=O)
C-C O-H 6(O-H)
5(C-H) 3(O=O)

3 CO2(g) + 4 H2O(l)
 –1015 + 4(C=O) + 6(O-H) = 44 + (C-C) + 5(C-H) + (C-O) + (O-H) + 3(O=O)
 (C-C) = –1015 + 4(C=O) + 6(O-H) - 44 - 5(C-H) - (C-O) - (O-H) - 3(O=O)
 (C-C) = –1015 + 4(743) + 6(463) - 44 - 5(412) - (360) - (463) - 3(498)
-1
= +314 kJ mol




PAGE 8 EXAMPLES
Ex 1 q = mcT
= 100 x 4.18 x 36 = 15048 J = 15.048 kJ

moles CH3CH2CH2OH = mass / Mr = 0..0 = 0.00833

H = q / moles = –15..00833
-1
= – 1810 kJ mol (3sf)




© www.CHEMSHEETS.co.uk 23-Feb-2016 Chemsheets A2 1014 Page 3

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