The amount of the heat released when 20 ml 0.5 M NaOH is mixed with 100 ml 0.1 M HCI is ´ kJ. The heat of neutralization is (A) –100 x kJ/mol (B) –50 x kJ/mol (C) +100 x kJ/mol (D) +50 x kJ/mol

To determine the heat of neutralization when mixing sodium hydroxide (NaOH) and hydrochloric acid (HCl), we first need to calculate the moles of each reactant involved in the reaction. The heat of neutralization is typically expressed in kJ/mol and represents the energy change when an acid and a base react to form water and a salt.

Step 1: Calculate Moles of Reactants

We start by calculating the number of moles of NaOH and HCl in the given solutions.

• For NaOH: The concentration is 0.5 M, and the volume is 20 ml (which is 0.020 L).
• Moles of NaOH: Moles = Concentration × Volume = 0.5 mol/L × 0.020 L = 0.01 moles.

• For HCl: The concentration is 0.1 M, and the volume is 100 ml (which is 0.100 L).
• Moles of HCl: Moles = Concentration × Volume = 0.1 mol/L × 0.100 L = 0.01 moles.

Step 2: Identify the Limiting Reactant

In this case, both NaOH and HCl are present in equal amounts (0.01 moles each). Therefore, neither is in excess, and they will completely react with each other.

Step 3: Heat of Neutralization Calculation

The heat of neutralization for strong acids and bases, like HCl and NaOH, is typically around -57 kJ/mol. This value indicates that the reaction is exothermic, meaning it releases heat.

Since we have 0.01 moles of each reactant, we can calculate the total heat released:

• Heat released: Heat = Moles × Heat of neutralization = 0.01 moles × (-57 kJ/mol) = -0.57 kJ.

Step 4: Expressing Heat of Neutralization

To express the heat of neutralization per mole, we can convert the total heat released into a per mole basis. Since we are looking for the heat of neutralization per mole of water formed, we can use the heat released for 1 mole of reaction:

• Heat of neutralization: -57 kJ/mol.

Final Answer

Now, looking at the options provided:

• (A) –100 x kJ/mol
• (B) –50 x kJ/mol
• (C) +100 x kJ/mol
• (D) +50 x kJ/mol

None of the options exactly match -57 kJ/mol, but if we consider the closest approximation, the heat of neutralization can be rounded to -50 kJ/mol, which corresponds to option (B). Therefore, the correct answer is: