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Standard XII

Chemistry

Enthalpy of Combustion

When 1.0g o...

Question

When $1.0 g$ of oxalic acid ( $H_{2} C_{2} O_{4}$ ) is burned in a bomb calorimeter whose heat capacity is $8.75 k J / K$ , the temperature increases by $0.312 K$ . The enthalpy of combustion of oxalic acid at $27^{o} C$ is :
$H_{2} C_{2} O_{4} (l) + \frac{1}{2} O_{2} (g) ⟶ H_{2} O (l) + 2 C O_{2} (g)$

- 245.7 k J / m o l

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- 244.452 k J / m o l

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- 246.947 k J / m o l

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none of these

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Solution

The correct option is A $- 245.7 k J / m o l$
As we know,
$Δ H = - C \times Δ T \times \frac{m o l a r m a s s}{m a s s t a k e n} = - 8.75 \times 0.312 \times \frac{90}{1} = - 245.7 k J / m o l$

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Similar questions

When $1.0 g$ of oxalic acid ( $H_{2} C_{2} O_{4})$ is burned in a bomb calorimeter whose heat capacity is $8.75 k J / K$ , the temperature increased by $0.312 K$ . The enthalpy of combustion of oxalic acid at $27^{\circ} C$ .

Q. When

1.0 g

of oxalic acid

(H_{2} C_{2} O_{4})

is burned in a bomb calorimeter whose heat capacity is

8.75 k J / K

, the temperature increases by

0.312 K

. The enthalpy of combustion of oxalic acid at

27^{o} C

is:

Q. When 1 g of anhydrous oxalic acid is burnt at

25^{0} C

, the amount of heat liberated is

2.835 k J

△ H

combustion is: (oxalic acid:

C_{2} H_{2} O_{4}

Q. Internal energy change for combustion of oxalic acid (

H_{2} C_{2} O_{4}

) is

- 245.7 k J / m o l

. The enthalpy of the combustion of oxalic acid at

27^{o}

C is:

Heat of combustion of benzoic acid $(C_{6} H_{5} C O O H)$ at constant volume at $25^{0}$ is 3233KJ/mole. When 0.5g of benzoic acid is burnt in bomb calorimeter, the temperature of calorimeter increased by ${0.53}^{0} C$ . Now in the same bomb calorimeter 1g of $C_{2} H_{6}$ burnt then temperature increased by ${2.04}^{0} C$ . $△ H$ for combustion of $C_{2} H_{6}$ is?

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When 1.0g of oxalic acid (H2C2O4) is burned in a bomb calorimeter whose heat capacity is 8.75kJ/K, the temperature increases by 0.312K. The enthalpy of combustion of oxalic acid at 27oC is :H2C2O4(l)+12O2(g)⟶H2O(l)+2CO2(g)

The correct option is A −245.7kJ/molAs we know,ΔH=−C×ΔT×molar massmass taken=−8.75×0.312×901=−245.7kJ/mol

The correct option is A $- 245.7 k J / m o l$
As we know,
$Δ H = - C \times Δ T \times \frac{m o l a r m a s s}{m a s s t a k e n} = - 8.75 \times 0.312 \times \frac{90}{1} = - 245.7 k J / m o l$