SATHEE: Chemical Thermodynamics Question 398

Chemical Thermodynamics Question 398

Question: For the reaction $2 C_{6} H_{5} C O_{2} H (s) + 15 O_{2} (g) \overset{}{\to}$

$14 C O_{2} (g) + 6 H_{2} O (g)$

$Δ U^{\circ} = - 772.7 k J m o l^{- 1}$ at 298 K. Calculate $Δ H^{\circ}$

Options:

A) $+ 760.3 k J m o l^{- 1}$

B) $- 760.3 k J m o l^{- 1}$

C) $+ 670.3 k J m o l^{- 1}$

D) $- 790.3 k J m o l^{- 1}$

Show Answer

Answer:

Correct Answer: B

Solution:

$Δ H^{^{\circ}} = Δ U^{^{\circ}} + Δ n R T$

$Δ H^{^{\circ}} = - 772.7 + \frac{5 \times 8.314 \times 298}{1000}$

$= - 760.3 k J, m o l^{- 1}$

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Chemical Thermodynamics Question 398

Question: For the reaction 2C6H5CO2H(s)+15O2(g)→

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Question: For the reaction $2 C_{6} H_{5} C O_{2} H (s) + 15 O_{2} (g) \overset{}{\to}$