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 $- 651 k c a l$ at $17^{0} C$ . Calculate the heat of reaction at constant volume at $17^{0} C$

- 654 k c a l

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- 327 k c a l

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- 664 k c a l

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+ 664 k c a l

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Solution

The correct option is A $- 654 k c a l$
$Δ H = Δ U + Δ n R T$
$Δ H = - 651$ x $10^{3} c a l$
$R = 2 c a l$
$T = 290 K$
$Δ n = (6 + 6) - 6 = 6$
$T h e r e f o r e, - 651$ x $10^{3} = Δ U + 6$ x $2$ x $290$
$Δ U = - 654480 c a l (o r) - 654.48 k c a l$

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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.

Q. 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 (g)

, at constant pressure is

- 651 K c a l

17^{o} C .

Calculate the heat of reaction at constant volume at

17^{o} 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 $17^{\circ} C$ ?

Q. The heat of reaction for

C_{10} H_{8} (s) + 12 O_{2} (g) \to 10 C O_{2} (g) + 4 H_{2} O (l)

at constant volume is (1229.892 Kcal at

25^{0}

C. Calculate the heat of reaction at constant pressure at

25^{0}

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Heat of reaction for, C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(v) at constant pressure is −651kcal at 170C. Calculate the heat of reaction at constant volume at 170C

The correct option is A −654kcalΔH = ΔU+ ΔnRT ΔH=−651x103 cal R =2 cal T =290K Δn =(6+6)−6 =6 Therefore, −651x103 = ΔU+6x2x290 ΔU =−654480 cal (or) −654.48 kcal

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 $- 651 k c a l$ at $17^{0} C$ . Calculate the heat of reaction at constant volume at $17^{0} C$

The correct option is A $- 654 k c a l$
$Δ H = Δ U + Δ n R T$
$Δ H = - 651$ x $10^{3} c a l$
$R = 2 c a l$
$T = 290 K$
$Δ n = (6 + 6) - 6 = 6$
$T h e r e f o r e, - 651$ x $10^{3} = Δ U + 6$ x $2$ x $290$
$Δ U = - 654480 c a l (o r) - 654.48 k c a l$