For the reactionAI - 2Bg -U- 2-1 kcal- -S- 20 calK-1at 300 K- Hence -G in kcal is

Step 1:∆H= ∆U + ∆n_gRTStep 2:∆H= 2100 + (2 × 2 × 300) (R=2 calK^ 1mol^ 1) 0ex0ex= 3300 calStep 3:∆G= ∆H – T∆S0ex0ex ∆G= 3300 – (300 × 20)0ex0ex = 2.7 kcal ...

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Strength of Materials

Double Integration -Cantilever - Standard Cases

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Question

For the reaction $A (I) \to 2 B (g) ∆ U = 2.1 kcal, ∆ S = 20 cal K^{- 1}$ at $300 K$ , Hence $∆ G$ in $kcal$ is

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Solution

Step 1:
$∆ H = ∆ U + ∆ n_{g} RT$
Step 2:
$∆ H = 2100 + (2 \times 2 \times 300) (R = 2 cal K^{- 1} {mol}^{- 1}) = 3300 cal$
Step 3:
$∆ G = ∆ H - T ∆ S ∆ G = 3300 - (300 \times 20) = - 2.7 kcal .$
Therefore, $∆ G$ is $- 2.7 kcal$ .

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For the reactionAI→2Bg∆U=2.1kcal,∆S=20calK-1at 300K, Hence ∆G in kcal is

Step 1:∆H=∆U+∆ngRTStep 2:∆H=2100+(2×2×300)(R=2calK-1mol-1)=3300calStep 3:∆G=∆H–T∆S∆G=3300–(300×20)=-2.7kcal.Therefore, ∆G is -2.7kcal.

For the reaction $A (I) \to 2 B (g) ∆ U = 2.1 kcal, ∆ S = 20 cal K^{- 1}$ at $300 K$ , Hence $∆ G$ in $kcal$ is

Step 1:
$∆ H = ∆ U + ∆ n_{g} RT$
Step 2:
$∆ H = 2100 + (2 \times 2 \times 300) (R = 2 cal K^{- 1} {mol}^{- 1}) = 3300 cal$
Step 3:
$∆ G = ∆ H - T ∆ S ∆ G = 3300 - (300 \times 20) = - 2.7 kcal .$
Therefore, $∆ G$ is $- 2.7 kcal$ .