13) l
since ccl4is liquid at its standard state
Q1. ΔUo of combustion of methane is -X kJ/mol. The value of ΔHo is
(a) =ΔUo (b) >ΔUo (c) <ΔUo (d) = 0
Q2. For an isolated system, ΔU=0. what will be ΔS ? (sign)
Q3. What is the equation for Heat of formation (standard) of CCl4 ?
C(grap,s) + 2Cl2(g) --> CCl4(l or g)??
Q4. NCERT QUES.
Calculate the enthalpy change on freezing of 1.0 mole of water at 10°C to ice at -10°C. ΔHfus = 6.03 kJ/mol at 0°C
Cp[H2O(l)] = 75.3 J/mol/K
Cp[H2O(s)] = 36.8 J/mol/K (my answer -7.151 kJ while ans given -6.415 kJ)
Q5. Calculate the number of kJ of heat necessary to raise the temp of 60.0 g of aluminium from 35°C to 55°C. molar heat capacity of Al is 24 J/mol/K
(my ans 1.067 kJ given 1.09 kJ)
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15 Answers
14 )
your answer. ans in book should be wrong.
5)
your ans. is correct.
106all questions are NCERT..
1. (c)
2. >0
tush, can u explain Q1. ??
2,3 still unsolved
11Ans 1) Use ΔH = ΔU + Δn RT
where Δn represents change in no. of moles of gaseous reactants and pdts.
Since Δn < 0 , therefore, ΔH < ΔU
42) " The entropy in an isolated system does not decrease. "
" A glass full of water might fall from a table to be broken into pieces with water scattered, but you cannot expect glass fragments, gathering water, will of themselves get restored to their original condition and jump onto the table. Entropy increases or at least never decreases. "
Source is Wikipedia
Well this is a fact
Logically we can prove it this way
ΔU = 0
ΔS = ΔH/T
ΔH > 0 ( since in an isolated system Δn ≥ 0 ---> use the abv example to understand y)
So ΔS ≥ 0
43) Ccl4 ( l )
"standard heat of formation" of a compound is the change of enthalpy that accompanies the formation of 1 mole of a substance in its standard state from its constituent elements in their standard states (the most stable form of the element at 1 bar of pressure and the specified temperature, usually 298.15 K or 25 degrees Celsius)
45) He must have considered at wt a bit less than 27
4) Did u get ?? If Yes Can u post the solution plzz
49yea the process involves 3 steps.....first water at 10°C colls to water at 0°C...then water at 0°C freezes to give ice at 0°C ,....then ice at 0°C cools to ice at -10°C..
so ΔH=ΔH(water cooling)+ΔH(water freezing)+ΔH(ice cooling)=nCPwaterdT+nΔHfusion+nCPicedT'
CP ice, CP water and ΔHfusion are given.
dT=-10°C
dT'=-10°C
n=1 mole..
so done!!![1][1][1]
106sorry Uttara..
Boards mein atak gaya tha
4th one i guess NCERT ans wrong...
working is how subhomoy has shown
thanks for 2,3,5
@all: comparing ΔH and ΔE, do we look at signs included or only magnitude???
1ΔH=ΔU+PΔV or ΔU+ΔngRT
Now , since its combustion , so work is done BY the system , it has to be -ve , and ΔU is given as -X
=> ΔH=-X+(-W) [where W=work)
=> ΔH less than U