when .250 mol of CH4... need explanation of answer please
when .250 mol of ch4 reacts with excess cl2 at constant pressure according to the chemical equation shown below. 177 kJ of heat ae released. calculate the value ofdelta H for this
2CH4(g) + 3 Cl2(g)---> 2CHCl3(l) + 3H2(g)
a)-1420 kJ
b)-708 kJ
c) +708 kJ
d) +1420 kJ
Homework Answers
Step1 Heat released by 2 moles of CH4 = 177x2/.25 =1416 kJ
Step2 ΔH= -1416 kJ /2=-1416/2=-708 kJ/mole
Species
Moles
CH4
0.250
Cl2
0.250*3/2=0.375
CHCl3
0.250*2/2=0.250
H2
0.250*3/2=0.375
ΔH=(HfCHCl3 *0.250)+(HfH2 *0.375)-(HfCH4 *0.250)-(HfCl2 *0.375)
Heat of formation(Hf ) for pure components is zero
ΔH=(-132*0.250)-(-74.8*0.250)=-14.3 kJ
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