Question

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

Answer 1

If you are expressing work done in terms of calories then you can use 2 Cal/mol K but if in joules, then … 8.314 J/mol K

2 used here, is nothing but an approximate value of

8.314/4.184

So, it comes from the relationship between J and Cal.

And please , R is not equal to 8.314 K J/mol K