Reaction Represented Chapter 6 Quiz Test

1. 4 NH3 (g) + 3 O2 (g) --> 2 N2 (g) + 6 H2O (g)If the standard molar heats of formation of ammonia, NH3 (g), and gaseous water, H2O (g), are -46 kJ/mol and -242 kJ/mol, respectively, what is the value of $capital delta H subscript 298 degree$ for the reaction represented above?

-190 kJ/mol

-290 kJ/mol

-580 kJ/mol

-1270 kJ/mol

-1640 kJ/mol

The value of ΔH for a reaction can be calculated using the standard molar heats of formation of the reactants and products. In this reaction, 4 moles of NH3 and 3 moles of O2 react to form 2 moles of N2 and 6 moles of H2O. The change in enthalpy is given by:

ΔH = (2 mol N2 * 0 kJ/mol) + (6 mol H2O * -242 kJ/mol) - (4 mol NH3 * -46 kJ/mol) - (3 mol O2 * 0 kJ/mol)
= -1270 kJ/mol

Therefore, the value of ΔH for the reaction is -1270 kJ/mol.

Explanation

The value of ΔH for a reaction can be calculated using the standard molar heats of formation of the reactants and products. In this reaction, 4 moles of NH3 and 3 moles of O2 react to form 2 moles of N2 and 6 moles of H2O. The change in enthalpy is given by:

ΔH = (2 mol N2 * 0 kJ/mol) + (6 mol H2O * -242 kJ/mol) - (4 mol NH3 * -46 kJ/mol) - (3 mol O2 * 0 kJ/mol)
= -1270 kJ/mol

Therefore, the value of ΔH for the reaction is -1270 kJ/mol.

2. $3 space C subscript 2 H subscript 2 left parenthesis g right parenthesis space rightwards arrow C subscript 6 H subscript 6 left parenthesis g right parenthesis$ What is the standard enthalpy change, $increment H degree$ , for the reaction represented above? ( $increment H degree subscript f$ for $C subscript 2 H subscript 2 left parenthesis g right parenthesis$ is 230 kJ/mol; $increment H degree subscript f$ of $C subscript 6 H subscript 6 left parenthesis g right parenthesis$ is 83 kJ/mol.)

-607 kJ

-147 kJ

-19 kJ

+19 kJ

+773 kJ

The standard enthalpy change for the reaction represented above is -607 kJ. This value is obtained by subtracting the enthalpy of the reactants (230 kJ/mol) from the enthalpy of the products (83 kJ/mol). The negative sign indicates that the reaction is exothermic, meaning it releases heat to the surroundings.

Explanation

The standard enthalpy change for the reaction represented above is -607 kJ. This value is obtained by subtracting the enthalpy of the reactants (230 kJ/mol) from the enthalpy of the products (83 kJ/mol). The negative sign indicates that the reaction is exothermic, meaning it releases heat to the surroundings.

3. A 6.22-kg piece of copper metal is heated from 20.5 degrees celsius to 324.3 degrees celsius. Calculate the heat absorbed (in kJ) by the metal. The specific heat of copper is 0.385 J/g*C

49.1 kJ

728 kJ

0.728 kJ

0.0491 kJ

777 kJ

The heat absorbed by an object can be calculated using the formula: Q = mcΔT, where Q is the heat absorbed, m is the mass of the object, c is the specific heat capacity of the material, and ΔT is the change in temperature. In this case, the mass of the copper is given as 6.22 kg, the specific heat capacity of copper is given as 0.385 J/g*C, and the change in temperature is (324.3 - 20.5) = 303.8 degrees Celsius. Converting the mass to grams, we get 6220 g. Plugging these values into the formula, we get Q = (6220 g)(0.385 J/g*C)(303.8 C) = 728 kJ. Therefore, the heat absorbed by the metal is 728 kJ.

Explanation

The heat absorbed by an object can be calculated using the formula: Q = mcΔT, where Q is the heat absorbed, m is the mass of the object, c is the specific heat capacity of the material, and ΔT is the change in temperature. In this case, the mass of the copper is given as 6.22 kg, the specific heat capacity of copper is given as 0.385 J/g*C, and the change in temperature is (324.3 - 20.5) = 303.8 degrees Celsius. Converting the mass to grams, we get 6220 g. Plugging these values into the formula, we get Q = (6220 g)(0.385 J/g*C)(303.8 C) = 728 kJ. Therefore, the heat absorbed by the metal is 728 kJ.

4. A 0.1375 g sample of solid magnesium is burned in a calorimeter that has a heat capacity of 3024 J/*C. If temperature increases by 1.126*C, calculate the enthalpy of the combustion of magnesium.

73.51 kJ/mol

602.0 J/mol

27.76 J/mol

602.0 kJ/mol

The enthalpy of the combustion of magnesium can be calculated using the formula:

Enthalpy change = (heat capacity of calorimeter * temperature change) / mass of magnesium

In this case, the heat capacity of the calorimeter is given as 3024 J/°C, the temperature change is given as 1.126°C, and the mass of magnesium is 0.1375 g. Plugging these values into the formula, we get:

Enthalpy change = (3024 J/°C * 1.126°C) / 0.1375 g

Simplifying the equation, we find:

Enthalpy change = 602.0 J/mol

Therefore, the enthalpy of the combustion of magnesium is 602.0 J/mol.

Explanation

The enthalpy of the combustion of magnesium can be calculated using the formula:

Enthalpy change = (heat capacity of calorimeter * temperature change) / mass of magnesium

In this case, the heat capacity of the calorimeter is given as 3024 J/°C, the temperature change is given as 1.126°C, and the mass of magnesium is 0.1375 g. Plugging these values into the formula, we get:

Enthalpy change = (3024 J/°C * 1.126°C) / 0.1375 g

Simplifying the equation, we find:

Enthalpy change = 602.0 J/mol

Therefore, the enthalpy of the combustion of magnesium is 602.0 J/mol.

5. $1 half H subscript 2 left parenthesis g right parenthesis space plus space 1 half I subscript 2 space left parenthesis s right parenthesis space rightwards arrow H I left parenthesis g right parenthesis space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space capital delta H equals 26 space k J divided by m o l$ $1 half H subscript 2 left parenthesis g right parenthesis space plus space 1 half I subscript 2 left parenthesis g right parenthesis space rightwards arrow H I left parenthesis g right parenthesis space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space capital delta H equals minus 5.0 space k J divided by m o l$ Based on the information above, what is the enthalpy change for the sublimation of iodine, represented below? $I subscript 2 left parenthesis s right parenthesis rightwards arrow I subscript 2 left parenthesis g right parenthesis$