Consider the decomposition of liquid benzene, C₆H₆(l), to gaseous acetylene, C₂H₂(g): C₆H₆(l) → 3 C2H₂(g) ΔH = +630 kJ (a) What is the enthalpy change for the reverse reaction?

Consider the decomposition of liquid benzene, C₆H₆(l), to gaseous acetylene, C₂H₂(g): C₆H₆(l) → 3 C2H₂(g) ΔH = +630 kJ (b) What is H for the formation of 1 mol of acetylene?

Consider the decomposition of liquid benzene, C₆H₆(l), to gaseous acetylene, C₂H₂(g): C₆H₆(l) → 3 C2H₂(g) ΔH = +630 kJ (c) Which is more likely to be thermodynamically favored, the forward reaction or the reverse reaction?

(b) The specific heat of aluminum is 0.9 J/(g - K). Calculate its molar heat capacity.

Two solid objects, A and B, are placed in boiling water and allowed to come to the temperature of the water. Each is then lifted out and placed in separate beakers containing 1000 g of water at 10.0 °C. Object A increases the water temperature by 3.50 °C; B increases the water temperature by 2.60 °C. (a) Which object has the larger heat capacity?

Two solid objects, A and B, are placed in boiling water and allowed to come to the temperature of the water. Each is then lifted out and placed in separate beakers containing 1000 g of water at 10.0 °C. Object A increases the water temperature by 3.50 °C; B increases the water temperature by 2.60 °C. (b) What can you say about the specific heats of A and B?

(a) What amount of heat (in joules) is required to raise the temperature of 1 g of water by 1 kelvin? (b) What amount of heat (in joules) is required to raise the temperature of 1 mole of water by 1 kelvin?

(d) How many kJ of heat are needed to raise the temperature of 5.00 kg of liquid water from 24.6 to 46.2 °C?

(b) Calculate the energy needed for this temperature change.

The specific heat of octane, C₈H₁₈(l), is 2.22 J•g/K. (a) How many J of heat are needed to raise the temperature of 80.0 g of octane from 10.0 to 25.0 °C?

The specific heat of octane, C₈H₁₈(l), is 2.22 J•g/K. (b) Which will require more heat, increasing the temperature of 1 mol of C₈H₁₈(l), by a certain amount or increasing the temperature of 1 mol of H₂O(l) by the same amount?

Consider the data about gold metal in Exercise 5.26(b). (a) Based on the data, calculate the specific heat of Au(s).

Consider the data about gold metal in Exercise 5.26(b). (b) Suppose that the same amount of heat is added to two 10.0-g blocks of metal, both initially at the same temperature. One block is gold metal, and one is iron metal. Which block will have the greater rise in temperature after the addition of the heat?

When a 6.50-g sample of solid sodium hydroxide dissolves in 100.0 g of water in a coffee-cup calorimeter (Figure 5.18), the temperature rises from 21.6 to 37.8 °C (a) Calculate the quantity of heat (in kJ) released in the reaction.

When a 6.50-g sample of solid sodium hydroxide dissolves in 100.0 g of water in a coffee-cup calorimeter (Figure 5.18), the temperature rises from 21.6 to 37.8 °C (b) Using your result from part (a), calculate H (in kJ/mol KOH) for the solution process. Assume that the specific heat of the solution is the same as that of pure water.

(b) Is this process endothermic or exothermic?

A 1.50-g sample of quinone (C₆H₄O₂) is burned in a bomb calorimeter whose total heat capacity is 8.500 kJ/°C. The temperature of the calorimeter increases from 25.00 to 29.49°C. (b) What is the heat of combustion per gram of quinone and per mole of quinone?

A 1.50-g sample of quinone (C₆H₄O₂) is burned in a bomb calorimeter whose total heat capacity is 8.500 kJ/°C. The temperature of the calorimeter increases from 25.00 to 29.49 °C. (a) Write a balanced chemical equation for the bomb calorimeter reaction.

A 2.20-g sample of phenol (C₆H₅OH) was burned in a bomb calorimeter whose total heat capacity is 11.90 kJ/°C. The temperature of the calorimeter plus contents increased from 21.50 to 27.50 °C. (a) Write a balanced chemical equation for the bomb calorimeter reaction.

A 2.20-g sample of phenol (C₆H₅OH) was burned in a bomb calorimeter whose total heat capacity is 11.90 kJ/°C. The temperature of the calorimeter plus contents increased from 21.50 to 27.50 °C. (b) What is the heat of combustion per mole of phenol?

Under constant-volume conditions, the heat of combustion of benzoic acid (C₆H₅O₆) is 15.57 kJ/g. A 3.500-g sample of sucrose is burned in a bomb calorimeter. The temperature of the calorimeter increases from 20.94 to 24.72 °C. (a) What is the total heat capacity of the calorimeter?

Under constant-volume conditions, the heat of combustion of benzoic acid (C₆H₅O₆) is 15.57 kJ/g. A 3.500-g sample of sucrose is burned in a bomb calorimeter. The temperature of the calorimeter increases from 20.94 to 24.72 °C. (b) If the size of the sucrose sample had been exactly twice as large, what would the temperature change of the calorimeter have been?

Under constant-volume conditions, the heat of combustion of naphthalene (C₁₀H₈) is 40.18 kJ/g. A 2.50-g sample of naphthalene is burned in a bomb calorimeter. The temperature of the calorimeter increases from 21.50 to 28.83 °C. (c) Suppose that in changing samples, a portion of the water in the calorimeter were lost. In what way, if any, would this change the heat capacity of the calorimeter?

Consider the following hypothetical reactions: A → B ΔH_I = +60 kJ B → C ΔH_II = -90 kJ (a) Use Hess’s law to calculate the enthalpy change for the reaction A → C.

Consider the following hypothetical reactions: A → B ΔH_I = +60 kJ B → C ΔH_II = -90 kJ (b) Construct an enthalpy diagram for substances A, B, and C, and show how Hess's law applies.

Calculate the enthalpy change for the reaction P₄O₆(s) + 2 O₂(g) → P₄O₁₀(s) given the following enthalpies of reaction: P₄(s) + 3 O₂(g) → P₄O₆(s) ΔH = -1640.1 kJ P₄(s) + 5 O₂(g) → P₄O₁₀(s) ΔH = -2940.1 kJ

From the enthalpies of reaction H₂(g) + F₂(g) → 2 HF(g) ΔH = -537 kJ C(s) + 2 F₂(g) → CF₄(g) ΔH = -680 kJ 2 C(s) + 2 H₂(g) → C₂H₄(g) ΔH = +52.3 kJ Calculate H for the reaction of ethylene with F₂: C₂H₄(g) + 6 F₂(g) → 2 CF₄(g) + 4 HF(g)

Given the data N₂(g) + O₂(g) → 2 NO(g) ΔH = +180.7 kJ 2 NO(g) + O₂(g) → 2 NO₂(g) ΔH = -113.1 kJ 2 N₂O(g) → 2 N₂(g) + O₂(g) ΔH = -163.2 kJ use Hess's law to calculate ΔH for the reaction N₂O(g) + NO₂(g) → 3 NO(g)

(a) Why does the standard enthalpy of formation of both the very reactive fluorine (F₂) and the almost inert gas nitrogen (N₂) both read zero?