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Why the value of R is 2 calories here? Question: Heat of reaction for C6H12O6(s) + 6O2 → 6CO2(g) + 6H2O(v)at constant pressure is −651 K cal at 17∘ C. What is the heat of reaction at constant volume at 17∘ C? Chemistry Chemical Thermodynamics Thermochemistry Solution: We know that Δ H = Δ U + Δ nRT Given Δ H = −651 × 103cal; R = 2 calorie T = 290K; Δ n = 6 + 6 − 6 = 6 ∴ −651 × 103 = Δ U + 6 × 2 × 290 Δ U = −654.48 kcal |
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Answer» Why the value of R is 2 calories here? Question: Heat of reaction for C6H12O6(s) + 6O2 → 6CO2(g) + 6H2O(v)at constant pressure is −651 K cal at 17∘ C. What is the heat of reaction at constant volume at 17∘ C? Chemistry Chemical Thermodynamics Thermochemistry Solution: We know that Δ H = Δ U + Δ nRT Given Δ H = −651 × 103cal; R = 2 calorie T = 290K; Δ n = 6 + 6 − 6 = 6 ∴ −651 × 103 = Δ U + 6 × 2 × 290 Δ U = −654.48 kcal |
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