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The change in enthalpy when 5 mole of liquid benzene transforms to the vapour state at its boiling temperature(90°C) and at 1 bar pressure is (ΔSV = 87J K -mol) _____________________KJ. (Round off to two decimal places).
Correct answer is between '157.80,158.00'. Can you explain this answer?
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To calculate the change in enthalpy when 5 moles of liquid benzene transforms to the vapor state at its boiling temperature (90°C), we need to use the enthalpy of vaporization.
The enthalpy of vaporization, ΔHvap, is the amount of heat energy required to convert one mole of a substance from the liquid phase to the gas phase at a constant temperature and pressure.
The enthalpy of vaporization of benzene is approximately 30.72 kJ/mol.
Therefore, the change in enthalpy, ΔH, can be calculated using the formula:
ΔH = ΔHvap x number of moles
ΔH = 30.72 kJ/mol x 5 moles
= 153.6 kJ
So, the change in enthalpy when 5 moles of liquid benzene transforms to the vapor state at its boiling temperature is 153.6 kJ.
The enthalpy of vaporization, ΔHvap, is the amount of heat energy required to convert one mole of a substance from the liquid phase to the gas phase at a constant temperature and pressure.
The enthalpy of vaporization of benzene is approximately 30.72 kJ/mol.
Therefore, the change in enthalpy, ΔH, can be calculated using the formula:
ΔH = ΔHvap x number of moles
ΔH = 30.72 kJ/mol x 5 moles
= 153.6 kJ
So, the change in enthalpy when 5 moles of liquid benzene transforms to the vapor state at its boiling temperature is 153.6 kJ.
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The change in enthalpy when 5 mole of liquid benzene transforms to the vapour state at its boilingtemperature(90°C) and at 1 bar pressure is(ΔSV = 87J K -mol) _____________________KJ.(Round off to two decimal places).Correct answer is between '157.80,158.00'. Can you explain this answer?
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