Question

1. Laboratory experiment, 30. J of heat was added to a gas piston assembly as the piston did 75 J of work on it surroundings. Calculate triangle E for the system

2. A 15.0 g piece of graphite is heated to 100.0°C and placed in a calorimeter. The graphite releases 815.1 J of heat to reach a final temperature of 23.9°C. What is the specific heat of graphite?

3. Calculate the standard enthalpy of reaction for the reaction 2 Na(s) + 2 H2O(l)----> 2 NaOH(aq) + H2(g). Standered enthalpies of formation are -285.8 kJ/Mom for H2O(l) and -470.11 kJ/mol for NaOH(aq)

Answer 1

Answer on the question #70170, Chemistry / Other

Question:

1. Laboratory experiment, 30. J of heat was added to a gas piston assembly as the piston did 75 J of work on it surroundings. Calculate triangle E for the system

2. A 15.0 g piece of graphite is heated to 100.0°C and placed in a calorimeter. The graphite releases 815.1 J of heat to reach a final temperature of 23.9°C. What is the specific heat of graphite?

3. Calculate the standard enthalpy of reaction for the reaction 2 Na(s) + 2 H2O(l)---> 2 NaOH(aq) + H2(g). Standered enthalpies of formation are -285.8 kJ/Mom for H2O(l) and -470.11 kJ/mol for NaOH(aq)

Answer:

1. The change in the energy of the system described equals heat transferred to the assembly minus the work made on its surroundings:

$$\Delta E = Q - W = 30J - 75J = -45J$$

2. The heat is related to the change in temperature of the closed system as:

$$Q = cm\Delta T;$$

Then, the specific heat is:

$$c = \frac{Q}{m\Delta T} = \frac{815.1\,J}{15.0\,g \cdot (100 - 23.9)^\circ C} = 0.714\,J\,g^{-1}\circ C^{-1}$$

3. According to the Hess law:

$$\begin{array}{l} \Delta H^\circ = 2\Delta H^\circ_f(NaOH) - 2\Delta H^\circ_f(H_2O) = -2 \cdot 470.11 + 2 \cdot 285.8 \\ = -368.6\,kJ\,mol^{-1} \end{array}$$

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