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Alex73 [517]
3 years ago
7

The bond enthalpy of the Br–Cl bond is equal to ΔH°rxn for the following reaction. BrCl(g) → Br(g) + Cl(g) Using the follo

wing data, what is the bond enthalpy of the Br–Cl bond?
a. Br2(l) → Br2(g) ΔH°rxn = 30.91 kJ/mol
b. Br2(g) → 2Br(g) ΔH°rxn = 192.9 kJ/mol
c. Cl2(g) → 2Cl(g) ΔH°rxn = 243.4 kJ/mol
d. Br2(l) + Cl2(g) → 2BrCl(g) ΔH°rxn = 29.2 kJ/mol
Chemistry
1 answer:
ddd [48]3 years ago
3 0

Answer:

a. Br2 (l) → Br2(g) ΔH°rxn = 30.91 kJ/mol

Explanation:

Bond enthalpy is the energy which is required to break 1 mol of of bonds in gaseous covalent molecules. Bond breaking process can be endothermic or exothermic depending on the molecules association. The process needs to be performed in chemical laboratory under standard conditions.

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A sample of He at 25C and 755 torr occupies a fixed volume of 16.8L. What mass of He must be pumped in to increase the pressure
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  • Volume (V): 16.8 L
  • Temperature (T): 25 °C
  • Final pressure (P₂): 1.87 atm

Step 2: Convert "P₁" to atm

We will use the conversion factor 1 atm = 760 torr.

755 torr × 1 atm/760 torr = 0.993 atm

Step 3: Convert "T" to K

We will use the following expression.

K = °C + 273.15

K = 25°C + 273.15 = 298 K

Step 4: Calculate the initial number of moles of He

We will use the ideal gas equation.

P₁ × V = n₁ × R × T

n₁ = P₁ × V/R × T

n₁ = 0.993 atm × 16.8 L/(0.0821 atm.L/mol.K) × 298 K

n₁ = 0.682 mol

Step 5: Calculate the final number of moles of He

We will use the ideal gas equation.

P₂ × V = n₂ × R × T

n₂ = P₂ × V/R × T

n₂ = 1.87 atm × 16.8 L/(0.0821 atm.L/mol.K) × 298 K

n₂ = 1.28 mol

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n = n₂ - n₁

n = 1.28 mol - 0.682 mol

n = 0.60 mol

Step 7: Convert "n" to mass

The molar mass of He is 4.00 g/mol

0.60 mol × 4.00 g/mol = 2.4 g

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