0.3 g of ethane undergoes combustion at 27°C in a bomb calorimeter. The temperature of calorimeter system (including the water) is found to rise by 0.5°C. The heat evolved during combustion of ethane at constant pressure is ________kJ mol–1.(Nearest integer)[Given : The heat capacity of the calorimeter system is 20 kJ K–1, R = 8.3 JK–1 mol–1.Assume ideal gas behaviour. Atomic mass of C and H are 12 and 1 g mol–1 respectively]
Question
0.3 g of ethane undergoes combustion at 27°C in a bomb calorimeter. The temperature of calorimeter system (including the water) is found to rise by 0.5°C. The heat evolved during combustion of ethane at constant pressure is ________kJ mol–1.(Nearest integer)[Given : The heat capacity of the calorimeter system is 20 kJ K–1, R = 8.3 JK–1 mol–1.Assume ideal gas behaviour. Atomic mass of C and H are 12 and 1 g mol–1 respectively]
Solution
To solve this problem, we need to follow these steps:
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First, we need to calculate the molar mass of ethane (C2H6). The atomic mass of carbon (C) is 12 g/mol and hydrogen (H) is 1 g/mol. So, the molar mass of ethane is (212) + (61) = 30 g/mol.
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Next, we need to calculate the number of moles of ethane. We know that the mass of ethane is 0.3 g. So, the number of moles of ethane is 0.3 g / 30 g/mol = 0.01 mol.
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Now, we can calculate the heat evolved during the combustion of ethane. We know that the temperature rise is 0.5°C and the heat capacity of the calorimeter system is 20 kJ/K. So, the heat evolved is 0.5°C * 20 kJ/K = 10 kJ.
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Finally, we need to calculate the heat evolved per mole of ethane. We know that the heat evolved is 10 kJ and the number of moles of ethane is 0.01 mol. So, the heat evolved per mole of ethane is 10 kJ / 0.01 mol = 1000 kJ/mol.
Therefore, the heat evolved during the combustion of ethane at constant pressure is approximately 1000 kJ/mol.
Similar Questions
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