In a constant volume calorimeter, `3.5 g` of a gas with molecular weight `28` was burnt in excess oxygen at `298.0 K`. The temperature of the calorimeter was found to increase from `298.0 K to 298.45 K` due to the combustion process. Given that the heat capacity of the calorimeter is `2.5 kJ K^(-1)`, find the numerical value for the enthalpy of combustion of the gas in `kJ mol^(-1)`
Text Solution
AI Generated Solution
To find the enthalpy of combustion of the gas in kJ/mol, we can follow these steps:
### Step 1: Calculate the number of moles of the gas
Given:
- Mass of the gas = 3.5 g
- Molecular weight of the gas = 28 g/mol
The number of moles (n) can be calculated using the formula:
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In a constant volume calorimeter, 3.5g of a gas with molecular weight 28 was burnt in excess oxygen at 298.0 K. The temperature of the calorimeter was found to increase from 298 K to 298 K due to the combustion process. Given that the combustion process. Given that the heat capacituy of the calorimeter is 2.5 J K^(-1) , the value for the enthalpy of combustion of the gas id
A
`90 kJ mol^(-1)`
B
`9 kJ mol^(-1)`
C
`18 kJ mol^(-1)`
D
`6 kJ mol^(-1)`
In a constant volume calorimeter 5g 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 kJ K^(-1),a numerical value for the DeltaU of combustion of the gas in kJ mol^(-1) is
