ΔH_c(CH₄)= - 890 kJ mol^-1
ΔH_c(C₂H₄) = -1423 kJ mol^-1
ΔH_c=-203.87 kJ mol^-1
Let the mixture contain x lit of CH₄ and (3.67 - x)
lit of ethylene
CH₄+2O₂\(\rightarrow \)2CO₂+2H₂O
XLit
C₂H₄+3O₂\(\rightarrow \)2CO₂+2H₂O
(3.67-X)Lit       2(3.67-X)Lit
Volume of Carbondioxide formed
=x + 2 (3.67 - x) = 6.11 lit
X+7.34-2X=6.11
7.34-X=6.11
X=1.23Lit
Given mixture contains 1.23 lit of methane and 2.44 lit of ethylene, hence
\(\triangle { H }_{ C }=\left[ \frac { \triangle { H }_{ C }\left( { CH }_{ 4 } \right) }{ 22.4Lit } \times \left( X \right) lit \right] +\left[ \frac { \triangle { H }_{ C }\left( { C }_{ 2 }{ H }_{ 4 } \right) }{ 22.4lit } \times \left( 3.67-X \right) lit \right] \)
\(\triangle { H }_{ C }=\left[ \frac { -890KJ{ mol }^{ -1 } }{ 22.4Lit } \times 1.23lit \right] +\left[ \frac { -1423 }{ 22.4lit } \times \left( 3.67-1.23 \right) lit \right] \)
ΔH_c =[-48.87kJ mol^-1]+[-155kJ mol^-1]
ΔH_c =-203.87kJ mol^-1 .

ΔH_f(Fe₂O₃)= -741 kJ mol^-1
ΔH_f(CO)= -137 kJ mol^-1
ΔH_f(CO₂)= -394.5 kJ mol^-1
Fe₂O₃ + 3CO ⟶ 2Fe + 3CO₂ ΔH_r=?
ΔH_r=Σ(ΔH_f)products - Σ(ΔH_f)reactants
ΔH_r=[2ΔH_f=(Fe)+3ΔH_f(CO₂)]-[ΔH_f(Fe₂O₃)+3ΔH_f(CO)]
ΔH_r=[0 + 3 (-394.5)] - [-741 +3 (-137)]
ΔH_r=[-1183.5] - [-1152]
ΔH_r=-1183.5 + 1152
ΔH_r=-31.5 kJ mol^-1

ΔH =30.56 kJ mol^-1
= 30560 J mol^-1
ΔS=6.66 \(\times\) 10^-3 kJK^-1 mol^-1
T=? at which ΔG=0
ΔG=ΔH-TΔS
0=ΔH-TΔS
T=\(\frac { \Delta H }{ \Delta S } \)
T=\(\frac { 30.56kJ\quad mol^{ -1 } }{ 6.66\times { 10 }^{ -3 }kJK^{ -1 }mol^{ -1 } } \)
T = 4589K
(i) At 4589K; ΔG= 0 the reaction is in equilibrium.
(ii) at temperature below 4598 K, ΔH>TΔS
ΔG=ΔH- TΔS > 0, the reaction in the forward direction, is non-spontaneous. In other words the reaction occurs in the backward direction.

C₃H₈+5O₂\(\rightarrow \)3CO₂+4H₂O
\(\triangle { H }_{ C }^{ 0 }=-2220.2KJ\quad mo{ l }^{ -1 }\)....(1)
C+O ₂\(\rightarrow \)3CO₂
\(\triangle { H }_{F}^{0 }=-393.5KJ\quad mo{ l }^{ -1 }\)....(2)
\({ H }_{ 2 }+\frac { 1 }{ 2 } { O }_{ 2 }\rightarrow { H }_{ 2 }O\)
\(\triangle { H }_{F}^{0 }=-285.8KJ\quad mo{ l }^{ -1 }\)...(3)
3C+4H₂\(\rightarrow \)C₃H₈
\(\triangle { H }_{ C }^{ 0 }\)=?
(2) X3 \(\Rightarrow \)3C+3O₂  \(\rightarrow \)3CO₂
\(\triangle { H }_{F}^{0 }= \) -1180.5 KJ ....(4)
(3)X4\(\Rightarrow \) 4H₂ +2O₂ \(\rightarrow \)4H₂O
\(\triangle { H }_{F}^{0 }= \) -1143.2KJ ....(5)
(4)+(5)-(1)\(\Rightarrow \) 3C+3O₂+4H₂+2O₂+3CO₂ +4H2O\(\rightarrow \)3CO₂+4H₂O+C₃H₈+5O₂
\(\triangle { H }_{F}^{0 }= \) -1180.5-1143.2-(-2220.2)KJ
3C+4H₂\(\rightarrow \)C₃H₈
\(\triangle { H }_{F}^{0 }= \) -103.5KJ
Standard heat of formation of propane is
\(\triangle { H }_{ R }^{ 0 }\left( { C }_{ 3 }{ H }_{ 8 } \right) =-103.5KJ\)

n = 2 moles
V_i=500 ml = 0.5 L
V_f = 2 L
T=25⁰C = 298 K
w=-2.303 nRTlog\((\frac{V_f}{V_i})\)
w=-2.303 \(\times\) 2 \(\times\) 8.314 \(\times\) 298 \(\times\) log\((\frac{2}{0.5})\)
w -2.303 \(\times\) 2 \(\times\) 8.314 \(\times\) 298 \(\times\) log(4)
w=-2.303 \(\times\) 2 \(\times\) 8.314 \(\times\) 298 \(\times\) 0.6021
w=-6871 J
w=-6.871 KJ.