150 ml of 0.5N nitric acid solution at 25.35oC was mixed with 150 ml of 0.5N sodium hydroxide solution at the same temperature. The final temperature was recorded to be 28.77oC. The heat of neutralization of nitric acid with sodium hydroxide is:

The temperature change ΔT = 28.77oC - 25.35oC = 3.42oC

To calculate the heat of neutralization, we need to consider the heat capacity of the solution. Let's assume the specific heat capacity of the solution is approximately equal to that of water, which is 1 cal/g·°C. The total volume of the solution is 150 ml + 150 ml = 300 ml. Assuming the density of the solution is approximately 1 g/ml, the mass of the solution is 300 g.

The heat absorbed by the solution (q) can be calculated using the formula:

q = mcΔT

where:

• q = heat absorbed (in calories)
• m = mass of the solution (in grams) = 300 g
• c = specific heat capacity of the solution (in cal/g·°C) = 1 cal/g·°C
• ΔT = change in temperature (in °C) = 3.42 °C

q = (300 g)(1 cal/g·°C)(3.42 °C) = 1026 cal

To convert this to kilocalories, we divide by 1000:

q = 1026 cal / 1000 cal/kcal = 1.026 kcal

Since the reaction is exothermic (the temperature increased), the heat of neutralization will be negative. The number of moles of nitric acid and sodium hydroxide are equal and can be calculated as:

Number of moles = Normality × Volume (in Liters) = 0.5 N × 0.150 L = 0.075 moles

The heat of neutralization per mole is:

Heat of neutralization per mole = q / number of moles = -1.026 kcal / 0.075 moles ≈ -13.68 kcal/mol

Therefore, the heat of neutralization of nitric acid with sodium hydroxide is approximately -13.68 kcal/mol. This corresponds to option C.