# The molar enthalpy of vaporization of hexane (CH14) is 28.9 kl/mol, and its normal boiling point is 68.73 C What is the vapor pressure [Solved] Answer : The correct option is, (b) 115 J/mol.K Explanation : Formula used : where, = change in entropy = change in enthalpy of vaporization = 40.5 kJ/mol = boiling point temperature = 352 K Now put all the given values in the above formula, we get: Therefore, the standard entropy of vaporization of ethanol at its boiling point is 115 J/mol.K

The molar enthalpy of vaporization in kJ/mol for butane = 24.265 kJ/mol. Explanation: Firstly, we need to calculate the no. of moles of 4.00 g of liquid butane CH: n = mass/molar mass = (4.0 g)/(58.12 g/mol) = 0.0688 mol. 0.0688 mol of butane requires a gain in enthalpy of 1.67 kJ to be evaporized. Know using cross multiplication: 0.0688 mol of butane to be vaporized requires 1.67 kJ. 1.0 mol of butane to be vaporized requires kJ. 1.0 mol of butane to be vaporized requires = (1.0 mol)(1.67 kJ)/(0.0688 mol) = 24.265 kJ. The molar enthalpy of vaporization in kJ/mol for butane = 24.265 kJ/mol.

When we are given the amount of substance present, and the molar enthalpy of vaporization, we may simply use the formula: H = n *H(vap) To find the enthalpy change occurring H = 2.15 * 71.8 The value ofH is 154.27 kJ.

The molar enthalpy of vaporization in kJ/mol for butane = 24.265 kJ/mol. Explanation: Firstly, we need to calculate the no. of moles of 4.00 g of liquid butane CH: n = mass/molar mass = (4.0 g)/(58.12 g/mol) = 0.0688 mol. 0.0688 mol of butane requires a gain in enthalpy of 1.67 kJ to be evaporized. Know using cross multiplication: 0.0688 mol of butane to be vaporized requires 1.67 kJ. 1.0 mol of butane to be vaporized requires kJ. 1.0 mol of butane to be vaporized requires = (1.0 mol)(1.67 kJ)/(0.0688 mol) = 24.265 kJ. The molar enthalpy of vaporization in kJ/mol for butane = 24.265 kJ/mol.

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The molar enthalpy of vaporization in kJ/mol for butane = 24.265 kJ/mol. Explanation: Firstly, we need to calculate the no. of moles of 4.00 g of liquid butane CH: n = mass/molar mass = (4.0 g)/(58.12 g/mol) = 0.0688 mol. 0.0688 mol of butane requires a gain in enthalpy of 1.67 kJ to be evaporized. Know using cross multiplication: 0.0688 mol of butane to be vaporized requires 1.67 kJ. 1.0 mol of butane to be vaporized requires kJ. 1.0 mol of butane to be vaporized requires = (1.0 mol)(1.67 kJ)/(0.0688 mol) = 24.265 kJ. The molar enthalpy of vaporization in kJ/mol for butane = 24.265 kJ/mol.

Answer : The correct option is, (b) 115 J/mol.K Explanation : Formula used : where, = change in entropy = change in enthalpy of vaporization = 40.5 kJ/mol = boiling point temperature = 352 K Now put all the given values in the above formula, we get: Therefore, the standard entropy of vaporization of ethanol at its boiling point is 115 J/mol.K

a) increase, S is positive b) S = 0.089 kJ/K Explanation: a) Entropy is a thermodynamic quantity which measures the degree of randomness of a system. Greater the disorder, greater will be the entropy. For different states of matter the entropy increases from solid to liquid to gas. In the given example, when Br2(l) boils it changes into the gas phase Br2(l) Br2(g) Thus, entropy will increase. i.e. S(product) > S(reactant) and S is positive. b) It is given that: Br2(l) Br2(g) H = 29.6 kJ/mol