SATHEE: Chemical Thermodynamics Question 321

Chemical Thermodynamics Question 321

Question: $C_{2} H_{6} (g) + 3.5 O_{2} (g) \to 2 C O_{2} (g) + 3 H_{2} O (g)$

$Δ S_{v a p} (H_{2} O, ℓ) = x_{1} c a l, K^{- 1}$ (boiling point = $T_{1}$ ) $Δ H_{f} (H_{2} O, ℓ) = x_{2}$

$Δ H_{f} (C_{2} O,) = x_{2}$

$Δ H_{f} (C_{2} H_{6},) = x_{4}$

Hence, $Δ H$ for the reaction is

Options:

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

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

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

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

Show Answer

Answer:

Correct Answer: B

Solution:

$H_{2} O (l) \to H_{2} O (g) Δ H_{v a p} = Δ S_{v a p} T_{B . P .} = x_{1} T_{1}$

$Δ H_{f} (H_{2} O, g) = Δ H_{f} (H_{2} O, 1) + Δ H_{v a p} = x_{2} + x_{1} T_{1}$

$Δ H_{r e a c t i o n} = 2 Δ H_{f}, (C O_{2}, g) + 3 Δ H_{f} (H_{2} O, g) - Δ H_{f} (C_{2} H_{6}, g)$

$= 2 x_{3} + 3 (x_{2} + x_{1} T_{1}) - x_{4} = 2 x_{3} + 3 x_{2} + 3 x_{1} T_{1} - x_{4}$

Chemical Thermodynamics Question 321

Question: $C_{2} H_{6} (g) + 3.5 O_{2} (g) \to 2 C O_{2} (g) + 3 H_{2} O (g)$

Options:

Answer:

Solution:

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Chemical Thermodynamics Question 321

Question: C2H6(g)+3.5O2(g)→2CO2(g)+3H2O(g)

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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Question: $C_{2} H_{6} (g) + 3.5 O_{2} (g) \to 2 C O_{2} (g) + 3 H_{2} O (g)$