Question

According to the following reaction, how much energy is evolved during the reaction of 2.50 L B2H6 and 5.65 L Cl2? (Both gases are initially at STP.) The molar mass of B2H6 is 27.67 g/mol.

b2h6(g) 6 cl2(g) → 2 bcl3(g) 6 hcl(g) δh°rxn = -1396
A) 352 kJ
B) 508 kJ
C) 215 kJ
D) 156 kJ
E) 58.7 kJ

Answer 1

To calculate the energy evolved during the reaction between 2.50 L B2H6 and 5.65 L Cl2, we need to determine the change in enthalpy (ΔH) using the ideal gas law and the stoichiometry of the reaction. By converting the given volumes to moles and scaling them to match the stoichiometry, we can calculate the energy evolved as approximately -77.1 kJ.

Step-by-step explanation:

In order to determine the energy evolved during the reaction, we need to calculate the change in enthalpy (ΔH) of the reaction. The balanced equation shows that 6 moles of B2H6 react with 6 moles of Cl2 to produce 2 moles of BCl3 and 6 moles of HCl. From the provided information, we know that the standard molar enthalpy change (ΔH°rxn) of the reaction is -1396 kJ. By converting the given volumes of gases to moles using the ideal gas law, we can calculate the amount of energy evolved during the reaction.

Using the ideal gas law, we can convert the given volumes of B2H6 and Cl2 to moles:

• B2H6 (2.50 L) = 2.50 L * (1 mol / 22.4 L) = 0.1116 mol
• Cl2 (5.65 L) = 5.65 L * (1 mol / 22.4 L) = 0.2522 mol

Since reaction stoichiometry tells us that 6 moles of B2H6 react with 6 moles of Cl2, we need to scale the moles of B2H6 and Cl2 to match the stoichiometry:

• B2H6: 0.1116 mol * (6 mol / 6 mol) = 0.1116 mol
• Cl2: 0.2522 mol * (6 mol / 6 mol) = 0.2522 mol

Now that we have the moles of B2H6 and Cl2, we can calculate the energy evolved using: ΔH = ΔH°rxn * (moles of B2H6 / 2 moles) = -1396 kJ * (0.1116 mol / 2 mol) = -77.1 kJ

Therefore, the amount of energy evolved during the reaction of 2.50 L B2H6 and 5.65 L Cl2 is approximately -77.1 kJ.