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