What is the enthalpy change for,
2H₂O₂(l)→2H₂O(l) if heat of formation of H₂O₂(l) and H₂O (l) are –188 and –286 kJ/mol respectively?

–196 kJ/mol
+ 948 kJ/mol
+ 196 kJ/mol
–948 kJ/mol

Solution

2H₂O₂(l)→2HO(l)+O₂(g)ΔH=?
=[(2×-286)+(0)-(2×-188)]
=[-572+376]=-196 kj/mole

The values of heat of formation of SO₂ and SO₃ are –298.2kJ and –98.2 kJ. The heat of formation of the reaction SO₂(1/ 2)O₂→SO₃ will be

–200 kJ
–356.2 kJ
+ 200 kJ
– 396.2 kJ

Solution

SO₂+¹⁄₂O₂→SO₃

ΔH=\(\Delta H_{F(SO_{3})}^{^{\circ}}-\Delta H_{F(SO_{2})}^{^{\circ}}\)

= –98.2 + 298.2= 200 kJ/mole

The values of ΔH for the combustion of ethane and acetylene are –341.1 and –310.0 k cal, respectively.The better fuel is :

ethane
acetylene
both
none of these

Solution

M.W.of ethane (C₂H₆) = 30 gm.
M.W. of acetylene(C₂H₂) = 26 gm.
Heat evolved per gm of ethane
= ³⁴¹⁄₃₀= 11.36 cal/gm.
Heat evolved per gm of acetylene
=³¹⁰⁄₂₆= 11.92 cal/gm
So, acetylene is better fuel.

Given that C+O₂→CO₂:ΔHº=-x kj
2 CO+O₂→2CO₂:ΔHº=-y kj
the enthalpy of formation of carbon monoxide will be

\(\frac{2x-y}{2}\)
\(\frac{2x-y}{2}\)
2x-y
y-2x

Solution

Given C+O₂→CO₂,ΔHº=-x kj....(i)
2CO₂→2CO+O₂......(ii)
or CO₂→CO+1/2CO₂,ΔHº=+ y/2 kJ...(iii)
By adding no. (i) and (iii) eq.

C+O₂+CO₂→CO₂+CO+¹⁄₂O₂

C+¹⁄₂O₂→CO,ΔHº-y/2-x=\(\frac{y-2x}{2}\)kj

A well stoppered thermos flask contains some ice cubes.This is an example of a

closed system
open system
isolated system
non-thermodynamic system

Solution

In isolated system neither exchange of matter nor exchange of energy is possible with surroundings.

For a chemical reaction the enthalpy and entropy change are –2.5 × 10³ cals and 7.4 cals deg^-1 respectively. At 25ºC the reaction is:

spontaneous
non-spontaneous
reversible
irreversible

Solution

ΔH=-2.5×10³ cals; ΔS=7.4 Cals deg^-1
Now,ΔG=ΔH-TΔS
= -2.5×10^-3-298×7.4=-4705.2=-ve

For the reaction
A(g)+2B(g)→2C(g)+3D(g)
the change of enthalpy at 27°C is 19 kcal. The value of ΔE is

21.2 kcal
17.8 kcal
18.4 kcal
20.6 kcal

Solution

A(g)+2B(g)→2C(g)+3D(g)
Δn=5-3=2
ΔH=ΔE+nRT or ΔE=ΔH-nRT
= 19 –2 × 2 × 10–3× 300 = 17.8 kcal

For a reaction Ag₂O(s)→2Ag+O, the value of ΔH =132.6 kJ, ΔS =66 JK^-1mol^-1 The free energy change for the reaction will be zero at which of the temperature :

2000 K
2009 K
2023 K
2029 K

Solution

ΔG=ΔH-TΔS
0 = 132.6 × 1000 – T × 66
\(T=\frac{132.6\times 1000}{66}=2000K\)

Two moles of an ideal gas is expanded isothermally and reversibly from 1 litre to 10 litre at 300 K. The enthalpy change(in kJ) for the process is

11.4 kJ
– 11.4 kJ
0 kJ
4.8 kJ

Solution

ΔH= nC_P ΔT solution; since ΔT= 0so, ΔH= 0

(ΔH –ΔU) for the formation of carbon monoxide (CO) from its elements at 298 K is(R = 8.314J K^-1mol^-1)

–2477.57 J mol^-1
2477.57 J mol^-1
–1238.78 J mol^-1
1238.78 J mol^-1

Solution

C(s)+¹⁄₂O₂(g)→CO(g)

[Δn=1-¹⁄₂=¹⁄₂]

ΔH –ΔU =ΔnRT=¹⁄₂×18.314×298=1238.78 J mol^-1

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