Heat of reaction for $C_{6} H_{12} O_{6} (s) + 6 O_{2} \to 6 C O_{2} (g) + 6 H_{2} O (v)$ at constant pressure is $- 651 K c a l$ at $17^{\circ} C$ . What is the heat of reaction at constant volume at $17^{\circ} C$ ?

-654.48 Kcal

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-655.48 Kcal

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-653.48 Kcal

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Solution

The correct option is A
-654.48 Kcal

We know that $Δ H = Δ U + Δ n R T$

Given $Δ H = - 651 \times 10^{3} c a l$ ; $R$ = $2 c a l o r i e$

$T = 290 K$ ; $Δ n = 6 + 6 - 6 = 6$

$∴ - 651 \times 10^{3} = Δ U + 6 \times 2 \times 290$

$Δ U = - 654.48 k c a l$

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Heat of reaction for $C_{6} H_{12} O_{6} (s) + 6 O_{2} (g) \to 6 C O_{2} (g) + 6 H_{2} O (v)$ at constant pressure is -651 kcal at 17°C. Calculate the heat of reaction at constant volume at 17°C.

Heat of reaction for $C_{6} H_{12} O_{6} (s) + 6 O_{2} \to 6 C O_{2} (g) + 6 H_{2} O (v)$ at constant pressure is $- 651 K c a l$ at $17^{\circ} C$ . What is the heat of reaction at constant volume at $17^{\circ} C$ ?

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

Q. Heat of reaction for: