The Bond Energy Quiz! Chemistry Trivia!

1. To calculate the amount of heat absorbed as a substance melts, which of the following information is not needed?

The mass of the substance

The change in temperature

The specific heat of the substance

The density of the sample

The density of the sample is not needed to calculate the amount of heat absorbed as a substance melts. The heat absorbed during the melting process depends on the mass of the substance, the change in temperature, and the specific heat of the substance. The density of the sample is not directly related to the heat absorbed during melting.

Explanation

The density of the sample is not needed to calculate the amount of heat absorbed as a substance melts. The heat absorbed during the melting process depends on the mass of the substance, the change in temperature, and the specific heat of the substance. The density of the sample is not directly related to the heat absorbed during melting.

2. Calculate the quantity of released heat of combustion of 5.76g of methane gas (CH₄) in an excess amount of oxygen gas at constant pressure CH_4(g)+2O_2(g)→ CO_2(g)+2H₂O_(l) , ∆H°= -890kj/mol ,(C=12,H=1)

320.4

-2472

-320.4

2472

The correct answer is -320.4. This is because the given reaction is the combustion of methane gas, which releases heat. The enthalpy change (∆H°) for this reaction is -890 kJ/mol. To calculate the quantity of released heat for a given amount of methane gas, we need to use the molar mass of methane (16 g/mol) and the given mass of methane (5.76 g). By using the equation q = ∆H° * (mass/molar mass), we can calculate the released heat. In this case, q = -890 kJ/mol * (5.76 g/16 g/mol) = -320.4 kJ. Therefore, the correct answer is -320.4.

Explanation

The correct answer is -320.4. This is because the given reaction is the combustion of methane gas, which releases heat. The enthalpy change (∆H°) for this reaction is -890 kJ/mol. To calculate the quantity of released heat for a given amount of methane gas, we need to use the molar mass of methane (16 g/mol) and the given mass of methane (5.76 g). By using the equation q = ∆H° * (mass/molar mass), we can calculate the released heat. In this case, q = -890 kJ/mol * (5.76 g/16 g/mol) = -320.4 kJ. Therefore, the correct answer is -320.4.

3. If one energy level diagram shows a much larger fall from the reactants to the products when compared with another energy level diagram what would these tell you?

Less exothermic

More energy is released

Endothermic reaction

If one energy level diagram shows a much larger fall from the reactants to the products when compared with another energy level diagram, this would indicate that more energy is being released in the reaction. This suggests that the reaction is more exothermic, as a larger fall in energy levels indicates a greater release of energy.

Explanation

If one energy level diagram shows a much larger fall from the reactants to the products when compared with another energy level diagram, this would indicate that more energy is being released in the reaction. This suggests that the reaction is more exothermic, as a larger fall in energy levels indicates a greater release of energy.

4. In the following reaction : CH_4(g)+Br_2(g)→CH₃Br_(g)+HBr_(g) Bond - energy Br-Br=193kj/mol C-Br =276kj/mol C-H =413kj/mol H-Br =366kj/mol Is these reaction exothermic or endothermic reaction ?

Exothermic reaction

Endothermic reaction

This reaction is exothermic because the energy released when forming the products (CH3Br and HBr) is greater than the energy required to break the bonds in the reactants (CH4 and Br2). The bond energy values provided indicate that breaking the C-H and Br-Br bonds requires more energy than is released when forming the C-Br and H-Br bonds. Therefore, the overall reaction releases energy, making it exothermic.

Explanation

This reaction is exothermic because the energy released when forming the products (CH3Br and HBr) is greater than the energy required to break the bonds in the reactants (CH4 and Br2). The bond energy values provided indicate that breaking the C-H and Br-Br bonds requires more energy than is released when forming the C-Br and H-Br bonds. Therefore, the overall reaction releases energy, making it exothermic.

5. ∆H°_f for the formation of the rust (Fe₂O₃) is -826 kj/mol, how much energy is involved in the formation of 5 gm of rust?

25.9 kj

66 kj

25.9 j

66 j

The question asks for the amount of energy involved in the formation of 5 grams of rust (Fe2O3), given that the ∆H°f for the formation of rust is -826 kJ/mol. To find the energy involved in the formation of 5 grams of rust, we need to convert grams to moles. The molar mass of Fe2O3 is 159.7 g/mol, so 5 grams is equal to 5/159.7 moles. Multiplying this by the ∆H°f (-826 kJ/mol), we get -25.9 kJ. Therefore, the correct answer is 25.9 kJ.

Explanation

The question asks for the amount of energy involved in the formation of 5 grams of rust (Fe2O3), given that the ∆H°f for the formation of rust is -826 kJ/mol. To find the energy involved in the formation of 5 grams of rust, we need to convert grams to moles. The molar mass of Fe2O3 is 159.7 g/mol, so 5 grams is equal to 5/159.7 moles. Multiplying this by the ∆H°f (-826 kJ/mol), we get -25.9 kJ. Therefore, the correct answer is 25.9 kJ.

6. Calculate the energy change required to produce 3 mol of glucose

Bond	Energy (kj/mol)
C-C C=C C-O C=O C-H O-H	347 602 358 799 413 464

-8148

12440

8148

-12440

The energy change required to produce 3 mol of glucose can be calculated by summing up the bond energies of all the bonds involved in the formation of glucose. The given bond energies are in kilojoules per mole (kJ/mol). By multiplying the bond energies by the number of each type of bond in glucose and adding them up, we can find the total energy change.

Explanation

The energy change required to produce 3 mol of glucose can be calculated by summing up the bond energies of all the bonds involved in the formation of glucose. The given bond energies are in kilojoules per mole (kJ/mol). By multiplying the bond energies by the number of each type of bond in glucose and adding them up, we can find the total energy change.

7. Match the following Calculate the enthalpy change from bond energies for each of these reaction

Bond	Energy	Bond	Energy
H-H H-F C-H C-O C≡O C-C C-Cl	432 565 413 358 1072 347 339	F-F O-H Cl-Cl C=C C=O C=O(forCO2(g)) O=O	154 467 239 614 745 799 495

CH4(g)+H2O(l)→ CO(g)+3H2(g) 1

2H2(g)+O2(g)→2H2O(g)

CH3COOH(g)+CH3OH(g)→ CH3COOCH3(g)+H2O(l)

H2(g)+F2(g)→2HF(g)

H2CCH2(g)+Cl2→ClH2CCH2Cl(g)

Submit

8. Calculate the molar enthalpy for water vapour from the following reaction CH_4(g)+H2O_(l)→CH₃OH_(g)+H_2(g)+78 , the molar enthalpy for CH₄ &CH₃OH are 75 kj/mol , 239 kj/mol respectively

86 kj/mol

239kj/mol

-239kj/mol

-86kj/mol

The molar enthalpy for water vapor can be calculated by subtracting the molar enthalpies of the reactants from the molar enthalpies of the products. In this reaction, the molar enthalpy of CH4 is 75 kj/mol and the molar enthalpy of CH3OH is 239 kj/mol. Since water vapor is a product in the reaction, we can subtract the molar enthalpies of the reactants (CH4 and H2O) from the molar enthalpies of the products (CH3OH and H2) to find the molar enthalpy of water vapor. Therefore, the correct answer is 239 kj/mol.

Explanation

The molar enthalpy for water vapor can be calculated by subtracting the molar enthalpies of the reactants from the molar enthalpies of the products. In this reaction, the molar enthalpy of CH4 is 75 kj/mol and the molar enthalpy of CH3OH is 239 kj/mol. Since water vapor is a product in the reaction, we can subtract the molar enthalpies of the reactants (CH4 and H2O) from the molar enthalpies of the products (CH3OH and H2) to find the molar enthalpy of water vapor. Therefore, the correct answer is 239 kj/mol.

9. Calculate ∆H for the reaction of SO₂ with O₂ , SO_2(g)+2O_2(g)→2SO_3(g), ∆H°_f(SO₂)= -296.8kj/mol , ∆H°_f(SO₃)=-395.7kj/mol

-98.9

197.8

None of them

-197.8

The reaction is exothermic because the formation of 2 moles of SO3 releases energy. The enthalpy change (∆H) can be calculated using the formula: ∆H = Σ∆Hf(products) - Σ∆Hf(reactants). Given that ∆H°f(SO2) = -296.8 kJ/mol and ∆H°f(SO3) = -395.7 kJ/mol, we can substitute these values into the formula. ∆H = (2*(-395.7 kJ/mol)) - (-296.8 kJ/mol) = -791.4 kJ/mol + 296.8 kJ/mol = -494.6 kJ/mol. Since the reaction releases energy, the value of ∆H is negative. The correct answer, -197.8 kJ/mol, is obtained by dividing -494.6 kJ/mol by 2, as the reaction involves the formation of 2 moles of SO3.

Explanation

The reaction is exothermic because the formation of 2 moles of SO3 releases energy. The enthalpy change (∆H) can be calculated using the formula: ∆H = Σ∆Hf(products) - Σ∆Hf(reactants). Given that ∆H°f(SO2) = -296.8 kJ/mol and ∆H°f(SO3) = -395.7 kJ/mol, we can substitute these values into the formula. ∆H = (2*(-395.7 kJ/mol)) - (-296.8 kJ/mol) = -791.4 kJ/mol + 296.8 kJ/mol = -494.6 kJ/mol. Since the reaction releases energy, the value of ∆H is negative. The correct answer, -197.8 kJ/mol, is obtained by dividing -494.6 kJ/mol by 2, as the reaction involves the formation of 2 moles of SO3.

10. When 80 gm of ammonium nitrate are dissolved in an amount of water to form 1 L of solution ,then temperature decrease from 20°C to 14°C , calculate the enthalpy for the reaction(N=14,O=16,H=1)

- 25080

2580

25080

None of them

The enthalpy change for a reaction can be calculated using the equation:

ΔH = q / n

Where ΔH is the enthalpy change, q is the heat absorbed or released by the reaction, and n is the number of moles of the substance involved in the reaction.

In this case, we know that 80g of ammonium nitrate is dissolved in water to form 1L of solution. Using the molar mass of ammonium nitrate (80g/mol), we can calculate the number of moles of ammonium nitrate present in the solution.

n = mass / molar mass
n = 80g / 80g/mol
n = 1 mol

We also know that the temperature decreases from 20°C to 14°C. The heat absorbed or released by the reaction can be calculated using the equation:

q = m * C * ΔT

Where q is the heat absorbed or released, m is the mass of the solution, C is the specific heat capacity of the solution, and ΔT is the change in temperature.

Since we have 1L of solution, we can assume that the mass of the solution is 1000g (since 1g = 1mL for water). The specific heat capacity of water is 4.18 J/g°C.

q = 1000g * 4.18 J/g°C * (14°C - 20°C)
q = -25080 J

Since the temperature decreases, the reaction is exothermic and releases heat. Therefore, the enthalpy change is -25080 J/mol, which is equivalent to -25080 kJ/mol.

Explanation

The enthalpy change for a reaction can be calculated using the equation:

ΔH = q / n

Where ΔH is the enthalpy change, q is the heat absorbed or released by the reaction, and n is the number of moles of the substance involved in the reaction.

In this case, we know that 80g of ammonium nitrate is dissolved in water to form 1L of solution. Using the molar mass of ammonium nitrate (80g/mol), we can calculate the number of moles of ammonium nitrate present in the solution.

n = mass / molar mass
n = 80g / 80g/mol
n = 1 mol

We also know that the temperature decreases from 20°C to 14°C. The heat absorbed or released by the reaction can be calculated using the equation:

q = m * C * ΔT

Where q is the heat absorbed or released, m is the mass of the solution, C is the specific heat capacity of the solution, and ΔT is the change in temperature.

Since we have 1L of solution, we can assume that the mass of the solution is 1000g (since 1g = 1mL for water). The specific heat capacity of water is 4.18 J/g°C.

q = 1000g * 4.18 J/g°C * (14°C - 20°C)
q = -25080 J

Since the temperature decreases, the reaction is exothermic and releases heat. Therefore, the enthalpy change is -25080 J/mol, which is equivalent to -25080 kJ/mol.