20cm3 of 0.500moldm-3 H2SO4 was consumed to neutralize 50cm3 of 0.400moldm-3 NaOH. The temperature of the system is increased by 36 degree C. The heat capacity of the calorie meter was 39 JK-1, and the heat capacity of the solution is 4.2JK-1. Find the enthalpy of neutralization for the reaction.
To find the enthalpy of neutralization for the reaction, we can use the equation:
q = m * c * ΔT
Where:
q is the heat transferred
m is the mass of the solution
c is the specific heat capacity of the solution
ΔT is the change in temperature
First, let's calculate the heat transferred from the reaction. Since the reaction between H2SO4 and NaOH is a neutralization reaction, we can assume that the heat released by the reaction is absorbed by the solution and the calorimeter.
The heat transferred from the reaction can be calculated using the equation:
q = m * c * ΔT
The mass of the solution can be calculated by multiplying the volume of the solution (in liters) by the molarity of the solution (in mol/L). In this case, we have 50 cm3 of a 0.400 mol/dm3 NaOH solution, which can be converted to 0.050 L * 0.400 mol/L = 0.020 mol.
The specific heat capacity of the solution is given as 4.2 J/K.
The change in temperature (ΔT) is given as 36 °C. However, to use this value in the equation, we need to convert it to Kelvin by adding 273.15. Therefore, ΔT = (36 °C + 273.15) K = 309.15 K.
Now, we can calculate the heat transferred from the reaction:
q = (0.020 mol H2SO4) * (4.2 J/K) * (309.15 K)
Simplifying this equation, we find:
q = 26.0028 J
Since the reaction is exothermic, the heat released by the reaction is absorbed by the solution and the calorimeter. Therefore, the heat transferred from the reaction is equal to the heat absorbed by the solution and the calorimeter:
q = q(solution) + q(calorimeter)
The heat capacity of the calorie meter is given as 39 J/K. Therefore:
q(calorimeter) = (39 J/K) * (ΔT)
Substituting the given ΔT into the equation, we find:
q(calorimeter) = (39 J/K) * (36 °C + 273.15 K)
Simplifying this equation, we find:
q(calorimeter) = 14998.35 J
Now, we can calculate the heat absorbed by the solution:
q(solution) = q - q(calorimeter)
Substituting the given q and q(calorimeter) into the equation, we find:
q(solution) = 26.0028 J - 14998.35 J
Simplifying this equation, we find:
q(solution) = -14972.34 J (negative sign indicates heat absorption)
Finally, we can calculate the enthalpy of neutralization using the equation:
ΔH = q(solution) / moles of reacting substance
In this case, the moles of reacting substance can be obtained from the balanced chemical equation for neutralization of H2SO4 and NaOH, which is:
H2SO4 + 2NaOH -> Na2SO4 + 2H2O
From the balanced equation, it can be seen that 1 mole of H2SO4 reacts with 2 moles of NaOH. Therefore, the mole ratio between H2SO4 and NaOH is 1:2.
Since 20 cm3 of a 0.500 mol/dm3 H2SO4 solution is consumed, the mole of H2SO4 used can be calculated as:
mole of H2SO4 = (0.020 L * 0.500 mol/L) / 1 = 0.010 mol
Now, we substitute the calculated values into the equation to find the enthalpy of neutralization:
ΔH = (-14972.34 J) / (0.010 mol)
Simplifying this equation, we find:
ΔH = -1497234 J/mol
Therefore, the enthalpy of neutralization for the reaction is -1497234 J/mol.