0.562 g of graphite is kept in a bomb calorimeter in excess of oxygen at 298 K and 1 atm and burnt according to the reaction
C + O₂(g)---> CO₂(g).
During the reaction the temperature rises from 298 K to 298.8 K. If the heat capacity of the magnitude calorimeter and its contents is 20.7 kJ/K what is the enthalpy change for the above reaction at 298 K and 1 atm?

Open in App

Solution

Dear Student,

$∆ H = ∆ U + ∆ n_{g} RT For the reaction, C + O_{2} (g) \to {CO}_{2} (g) ∆ n_{g} = 1 - 1 = 0 Therefore, ∆ H = ∆ U = q_{V} ∆ H = q_{V} = C_{V} ∆ T = 20.7 {kJK}^{- 1} \times 0.8 K = 18.4 kJ As the reaction is a combustion reaction i . e . exothermic Therefore, ∆ H = - 18.4 kJ for the combustion of 0.562 g of C (graphite) For 1 mol C (graphite), ∆ H = - 18.4 \times \frac{12}{0.562} = - 392.88 {kJmol}^{- 1}$

Suggest Corrections

Similar questions

Q. 1 g of graphite is burnt in a bomb calorimeter in excess of oxygen at 298 K and 1 atmospheric pressure according to the equation

C (g r a p h i t e) + O_{2} (g) \to {C O}_{2} (g)

During the reaction, temperature rises from 298 K to 299 K. If the heat capacity of the bomb calorimeter is 20.7 kJ/K, what is the enthalpy change for the above reaction at 298 K and 1 atm?

Q. A

1.250 g

sample of octane

(C_{8} H_{18})

is burned in excess of oxygen in a bomb calorimeter. The temperature of calorimeter rises from

294.05 K

300.78 K

. If heat capacity of calorimeter and its contents is

8.93 k J / K

, calculate the heat transferred to calorimeter.

In a constant volume calorimeter, $3.5 g$ of a gas with molecular weight 28 g/mol was burnt in excess of oxygen at $298 K$ . The temperature of the calorimeter was found to Increase from $298 K$ to $298.45 K$ due to the combustion process. Given that the heat capacity of the calorimeter is $2.5 k J / K$ . The Numerical value for the enthalpy of combustion of the gas in kJ/mol is

Q. In a constant volume calorimeter,

5

g of a gas with molecular weight

40

was burnt in excess of oxygen at

298

K. The temperature of the calorimeter was found to increase from

298

K to

298.75

K due to combustion process. Given that the heat capacity of the calorimeter is

2.5

K^{- 1}

, the numerical value for the

Δ U

of combustion of the gas in kJ

m o l^{- 1}

is:

In a constant volume calorimeter, $3.5 g$ of a gas with molecular weight 28 was burnt in excess of oxygen at $298 k$ . The temperature of the calorimeter was found to Increase from $298 k$ to $298.45 k$ due to the combustion process. Given that the heat capacity of the calorimeter is $2.5 K J / K$ . The Numerical value for the enthalpy of combustion of the Gas in KJ /Mole is

Join BYJU'S Learning Program