C2H6(g) + 3.5 O2(g) →2CO2(g) + 3H2O(g) Δ S textvap(H2O,l)=x1cal K-1( textboiling point=T1) Δ Hf(H2O,l)=x2 Δ Hf(CO2)=x3 Δ Hf(C2H6)=x4 Hence, Δ H for the reaction is

Question

C2H6(g) + 3.5 O2(g) →2CO2(g) + 3H2O(g) Δ S textvap(H2O,l)=x1cal K-1( textboiling point=T1) Δ Hf(H2O,l)=x2 Δ Hf(CO2)=x3 Δ Hf(C2H6)=x4 Hence, Δ H for the reaction is (A) 2x3 + 3x2 - x4 (B) 2x3 + 3x2 - x4 + 3x1T1 (C) 2x++ 3x2x4-3x1 T1 (D) x1T1 +x2 + x3- x4

Tardigrade Admin · Accepted Answer

H2O(l)→ H2O(g)Δ H textvap=Δ S textvapTB.P=x1T1 Δ Hf(H2O,g)=Δ Hf(H2O,l)+Δ H textvap=x2+x1T1 Δ H textreaction=2Δ Hf(CO2,g)+3Δ Hf(H2Og)-Δ Hf(C2H6,g) =2x3+3(x2+x1T1)-x4 =2x3+3x2+3x1T1-x4

Q. $C_{2} H_{6} (g) + 3.5 O_{2} (g) \to 2 C O_{2} (g) + 3 H_{2} O (g)$
$Δ S_{vap} (H_{2} O, l) = x_{1} c a l K^{- 1} (boiling point = T_{1})$
$Δ H_{f} (H_{2} O, l) = x_{2}$
$Δ H_{f} (C O_{2}) = x_{3}$
$Δ H_{f} (C_{2} H_{6}) = x_{4}$
Hence, $Δ H$ for the reaction is

$2 x_{3} + 3 x_{2} - x_{4}$

$2 x_{3} + 3 x_{2} - x_{4} + 3 x_{1} T_{1}$

$2 x^{+} + 3 x_{2} x_{4} - 3 x_{1} T_{1}$

$x_{1} T_{1} + x_{2} + x_{3} - x_{4}$

Q. C2​H6​(g)+3.5O2​(g)→2CO2​(g)+3H2​O(g) ΔSvap​(H2​O,l)=x1​calK−1(boiling point=T1​) ΔHf​(H2​O,l)=x2​ ΔHf​(CO2​)=x3​ ΔHf​(C2​H6​)=x4​ Hence, ΔH for the reaction is

2x3​+3x2​−x4​

2x3​+3x2​−x4​+3x1​T1​

2x++3x2​x4​−3x1​T1​

x1​T1​+x2​+x3​−x4​

H2​O(l)→H2​O(g)ΔHvap​=ΔSvap​TB.P​=x1​T1​ ΔHf​(H2​O,g)=ΔHf​(H2​O,l)+ΔHvap​=x2​+x1​T1​ ΔHreaction​=2ΔHf​(CO2,g​)+3ΔHf​(H2​Og)−ΔHf​(C2​H6,​g) =2x3​+3(x2​+x1​T1​)−x4​=2x3​+3x2​+3x1​T1​−x4​

Q. $C_{2} H_{6} (g) + 3.5 O_{2} (g) \to 2 C O_{2} (g) + 3 H_{2} O (g)$
$Δ S_{vap} (H_{2} O, l) = x_{1} c a l K^{- 1} (boiling point = T_{1})$
$Δ H_{f} (H_{2} O, l) = x_{2}$
$Δ H_{f} (C O_{2}) = x_{3}$
$Δ H_{f} (C_{2} H_{6}) = x_{4}$
Hence, $Δ H$ for the reaction is

$2 x_{3} + 3 x_{2} - x_{4}$

$2 x_{3} + 3 x_{2} - x_{4} + 3 x_{1} T_{1}$

$2 x^{+} + 3 x_{2} x_{4} - 3 x_{1} T_{1}$

$x_{1} T_{1} + x_{2} + x_{3} - x_{4}$