The standard Gibbs energy change for the formation of propane C3 H8 g at 298 K is- -Given - H0f of propane is -103-85 kJ-mol S0m C3 H8g - 270-0 JK-2 mol-1 S0m H2g - 1309-68 JK-2 mol-1 S0m C- graphite - 5-79 JK-1 mol-1-

The correct option is A 1382 kJ / K /mol 3C(graphite)+4H_2(g) #160; → C_3H_8 (g) Δ S^0_f = S^0 (C3H8) #160; [3S^0 (C(graphite) #160; + 4S^ ...

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Standard X

Chemistry

Mole Concept and Avogadro's Number

The standard ...

Question

The standard Gibbs energy change for the formation of propane $C_{3} H_{8} g$ at 298 K is:

[Given $Δ H_{f}^{0}$ of propane is -103.85 kJ/mol
$S_{m}^{0} C_{3} H_{8} (g) = 270.0 J K^{- 2} m o l^{- 1}$
$S_{m}^{0} H_{2} (g) = 1309.68 J K^{- 2} m o l^{- 1}$
$S_{m}^{0} C$ : graphite $= 5.79 J K^{- 1} m o l^{- 1}$ ]

1382 kJ / K /mol

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-10.98

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12.354 kcal

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10.98 kcal

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Solution

The correct option is A 1382 kJ / K /mol
$3 C (g r a p h i t e) + 4 H_{2} (g) \to C_{3} H_{8} (g)$

$Δ S_{f}^{0} = S^{0} (C 3 H 8) - [3 S^{0} (C (g r a p h i t e) + 4 S^{0} (H 2)]$

$Δ S_{f}^{0} = 270.0 J / K / m o l - [3 (5.79 J / K / m o l) + 4 (1309.68 J / K / m o l)] = - 4986 J / K / m o l$
$Δ G_{f}^{0} = Δ H_{f}^{0} - T Δ S_{f}^{0}$

$Δ G_{f}^{0} = - 103850 J / m o l - 298 \times (- 4986 J / K / m o l)$

$Δ G_{f}^{0} = 1382005 J / K / m o l$

$Δ G_{f}^{0} = 1382 k J / K / m o l$

Hence, the correct option is $A$

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Similar questions

The standard Gibbs energy change for the formation of propane $(C_{3} H_{8 (g)})$ at 298 K is

$[Δ_{f} H^{θ}$ for propane $= - 103.85$ kJ mole $^{- 1},$ $S_{m}^{θ} C_{3} H_{8 (g)} = 270.2$ JK $^{- 1}$ mol $^{- 1}$ , $S_{m}^{θ} H_{2 (g)} = 130.68$ JK $^{- 1}$ mol $^{- 1}$ , $S_{m}^{θ} C_{(g r a p h i t e)} = 5.742$ JK $^{- 1}$ mol $^{- 1}]$

Q. Standard enthalpy and standard entropy changes for the oxidation of ammonia at 298 K are382.64 kJ mol

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Q. For the reaction given below the values of standard Gibbs free energy of formation at 298 K are given. What is the nature of reaction?

I_{2} + H_{2} S \to 2 H I + S

△ G_{f}^{o} (H I) = 1.8 k J m o l^{- 1}, △ G_{f}^{o} (H_{2} S) = 33.8 k J m o l^{- 1}

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Standard enthalpy and standard entropy for the oxidation of $N H_{3}$ at $298$ K are $- 382.64$ $kJ {mol}^{- 1}$ and $- 145.6$ $J {mol}^{- 1} K^{- 1}$ respectively. Standard free energy change (in ${kJ mol}^{- 1}$ ) for the same reaction at $298$ K is:

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The standard Gibbs energy change for the formation of propane C3H8g at 298 K is: [Given ΔH0f of propane is -103.85 kJ/molS0mC3H8(g)=270.0JK−2mol−1S0mH2(g)=1309.68JK−2mol−1S0mC: graphite =5.79JK−1mol−1]