Question

The following information is given for lead at latm: AH yap(1740 °C) -178 kJ/mol AH fus(328 °C) = 4.77 kJ/mol boiling point = 1740 °C melting point = 328 °C specific heat solid= 0.130 J/gºC specific heat liquid = 0.138 J/gºC What is AH in kJ for the process of freezing a 26.3 g sample of liquid lead at its normal melting point of 328 °C.

Answer 1

1) We have to calculate $\Delta$ H for converting 26.3 g of liquid lead to solid at 328 $\tiny \bullet$ C.

We have the $\Delta$ Hf ( 328 $\tiny \bullet$ C) = 4.77 kJ/mol, i.e this much enery is required to convert 1 mol of solid lead to liquid at given temp of 328 $\tiny \bullet$ C.

Now we know that molar mass of Lead = 207.2 u

Therefore moles of lead = 0.127

Now $\Delta$ H for the process = 0.127 * - 4.77 kJ = -0.6 kJ

The -ve sign indicates that heat is released from the system.

The same process above can be follwed for the next 2 ques by changing the appropriate numerical values.