Chemical Thermodynamics(12th Grade > Chemistry ) Questions and Answers
Explanation:-
Answer: Option A. -> +99.45 kcal:
A
The equation for bond dissociation can be given by
CH4(g) ⟶ C(g) + 4H(g)ΔH = 397.8 kcal
There are in total of 4 C - H bonds
Therefore the energy needed to break a C - H bond can be given by 397.84= + 99.45 kcal
Explanation:-
Answer: Option C. -> ΔSv=109.38 JK−1 mol−1:
C
For entropy change of vaporization can be given as, ΔSv=ΔHvT
Given values are, ΔHv=40.8×103Jmol−1; T=373K
Therefore ΔSv=(40.8×103373)
ΔSv=109.38JK−1mol−1
Explanation:-
Answer: Option D. -> Eb:
D
Enthalpy of reaction(ΔH)=Ea(f)−Ea(b) for an endothermic reactionΔH = + ve hencefor ΔH to be negativeEa(b) < Ea(f)
Explanation:-
Answer: Option C. -> 185 kJ:
C
Volume of C2H4 = 3.67×23=2.45Lit
Volume of CH4 = 1.22Lit
Mole of C2H5 = PV = nRTn = PVRT
= 1×2.450.082×298=n=0.1mole
Mole of CH4 = nCH4 = 1×1.220.082×298=0.05mole
∵ Energy released due to combusting 1 mole C2H4=1400KJ
∴ Energy released due to combusting 1 mole C2H4 = 1400×0.1=140kJ
Similar energy released in combusting 0.05 mole
CH4=−900×0.05=45
So that heat realized = 140+45 = 185kJ/mole
Explanation:-
Answer: Option D. -> 463 K, non-spontaneous:
D
Now, to get T, let's use
△G = △H−T△S
Here, △H=+30.558KJ
△S=0.066kJk−1
△G=0
⇒0=30.558−T×0.066
⇒T=30.5580.066=463K
Now, at 463 K⇒△G = 0
If we go below 463 K, T△S value will decrease and △G value will become positive.
Therefore, non-spontaneous
Explanation:-
Answer: Option B. -> Adiabatic process:
B
What do you mean by iso-entropic process?
⇒The name itself is giving you the answer.
The entropy should be same before and after the process occurs.
Here we know that,
ΔS=dqT
In case of adiabatic process,
dq = 0
Therefore ΔS = 0
Explanation:-
Answer: Option D. -> – 103 kJ:
D
ΔH∘=∑B.E(reactants)−∑B.E(products)=[B.E(H−H)+B.E(Br−Br)]−[2B.E(H−Br)]=(433+192)−(2×364)=625−728=−103kJ
Explanation:-
Answer: Option C. -> 104 Kcal:
C
Energy needed to dissociate 4g is 208 Kcal
∴ Number of moles of H2 = 2 mole
∴ Energy needed to dissociate 1 mole = 2082= 104 Kcal
∴ The bond energy is 104 Kcal in case of H - H bond.
Explanation:-
Answer: Option B. -> 1.0412 Pa:
B
W=−2.303nRTlogP1P2∴logP1P2=W−2.303nRT=100−2.303×1×8.314×298=−0.01753i.e.,logP1−logP2=−0.01753(TheunitsinwhichP1andP2areexpressedmustbeidentical)∴−logP2=−0.01753−logP1∴−logP2=−0.01753+logP1=0.01753+log1=0.01753+0logP2=0.01753⇒P2=Antilog(0.01753)⇒P2=1.0412Pa(SinceunitofP1isPa)