Question

A.) The standard enthalpy change for the following reaction is 359 kJ at 298 K.

PbCl₂(s) ----> Pb(s) + Cl₂(g)   ΔH° = 359 kJ

What is the standard enthalpy change for this reaction at 298 K?

Pb(s) + Cl2(g) ----> PbCl2(s)

__________ kJ

B.) The standard enthalpy change for the following reaction is -602 kJ at 298 K.

Mg(s) + 1/2 O2(g) ---->  MgO(s)   ΔH° = -602 kJ

What is the standard enthalpy change for the reaction at 298 K?

2 Mg(s) + O₂(g) ----> 2 MgO(s)

__________ KJ

C.) In the laboratory a "coffee cup" calorimeter, or constant pressure calorimeter, is frequently used to determine the specific heat of a solid, or to measure the energy of a solution phase reaction. A student heats 62.89 grams of tin to 98.70 °C and then drops it into a cup containing 79.80 grams of water at 21.69 °C. She measures the final temperature to be 24.76 °C. The heat capacity of the calorimeter (sometimes referred to as the calorimeter constant) was determined in a separate experiment to be 1.85 J/°C. Assuming that no heat is lost to the surroundings calculate the specific heat of tin.

Specific Heat (Sn) = __________J/g°C.

A chunk of nickel weighing 19.78 grams and originally at 97.27 °C is dropped into an insulated cup containing 83.54 grams of water at 22.22 °C. The heat capacity of the calorimeter (sometimes referred to as the calorimeter constant) was determined in a separate experiment to be 1.57 J/°C. Using the accepted value for the specific heat of nickel (0.444 C_sp (J/g^oC)), calculate the final temperature of the water. Assume that no heat is lost to the surroundings.

T_final =__________ °C.

Answer 1

A. In a reverse reaction the sign of enthalpy changes. It is because enthalpy is a state function meaning it has a unique value for a given state. For any given reaction enthalpy change for a forward reaction and that for a reverse reaction are equal in magnitude but opposite in phase.

Hence = -359kJ

B. When a thermodynamic equation is multiplied by a constant the thermodynamic quantity is also multiplied by same constant.

Hence    = 2*-602 = -1204kJ

C.

q = heat absorbed/released, s = specific heat, m = mass of substance, T = temperature

Heat released by Tin = Heat gained by water + Heat gained by calorimeter

62.89 *s* (98.70-24.76) = 79.80 *4.186* (24.76-21.69) + 1.85*(24.76-21.69)

[Note- specific heat of water is 1 cal/g^oC = 4.186J/g^oC]

4650.0866*s = 1025.511 + 5.6795

s = 0.2217J/g^oC

So, specific heat of tin is 0.2217J/g^oC

D.

Heat released by Nickel = Heat gained by water + Heat gained by calorimeter

19.78 *0.444* (97.27- T) = 83.54 *4.186* ( T- 22.22) + 1.85*( T -22.22)

854.256 - 8.7823T = 349.698T - 7770.299 + 1.85T - 41.107

8665.662 = 360.33T

T = 24.049^oC