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.

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Solution

As we know,
$△ H = △ E + △ n R T$
Here, $△ H = - 651 \times 10^{3}$ cal
$R = 2 c a l$
$T = 290 K$
and $△ n = 6 + 6 - 6$
$= 6$
$∴ - 651 \times 10^{3} = △ E + 6 \times 2 \times 290$
$△ E = - 651000 - 3480$
$= - 654.480$
$∴ △ E = - 654.48 K c a l$

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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. 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

17^{0} C

. Calculate the heat of reaction at constant volume at

17^{0} 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

Q. The heat of reaction

C (s) + \frac{1}{2} O_{2} (g) ⟶ C O (g)

17^{o} C

and at constant volume is

- 29.29 k c a l

. Calculate the heat of reaction at constant pressure.

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Heat of reaction for C6H12O6(s)+6O2(g)→6CO2(g)+6H2O(v) at constant pressure is -651 kcal at 17°C. Calculate the heat of reaction at constant volume at 17°C.

As we know,△H=△E+△nRTHere, △H=−651×103 calR=2calT=290Kand △n=6+6−6=6∴−651×103=△E+6×2×290△E=−651000−3480=−654.480∴△E=−654.48Kcal

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.

As we know,
$△ H = △ E + △ n R T$
Here, $△ H = - 651 \times 10^{3}$ cal
$R = 2 c a l$
$T = 290 K$
and $△ n = 6 + 6 - 6$
$= 6$
$∴ - 651 \times 10^{3} = △ E + 6 \times 2 \times 290$
$△ E = - 651000 - 3480$
$= - 654.480$
$∴ △ E = - 654.48 K c a l$