Calorimetry: Calculate Molar Enthalpy of Neutralization for Sulfuric Acid

[Physics197]

Homework Statement

A student mixed 200.00ml of 1.5mol/L sulphuric acid with 400.00ml of 1.5mol/L sodium hydroxide. Both solutions were at 19.67 degrees celsius initially and the highest temperature reached by the reaction mixture was 34.06 degrees celsius. Calculate the molar enthalpy of neutralization for sulphuric acid.

Homework Equations

q=nH
q=mc(delta)T

The Attempt at a Solution

Just need help with starting how to solve this problem.

 

[nate808]

Hello, to solve this problem, we can use the equation q=nH, where q is the heat released or absorbed by the reaction, n is the number of moles involved in the reaction, and H is the molar enthalpy of neutralization. To find n, we can use the equation n=cV, where c is the concentration and V is the volume. We know the concentrations of both solutions and their volumes, so we can calculate the number of moles for both sulphuric acid and sodium hydroxide.

Next, we need to calculate the heat released by the reaction using the equation q=mc(delta)T, where m is the mass of the solution, c is the specific heat capacity of the solution, and (delta)T is the change in temperature. We know the initial and final temperatures of the reaction mixture, so we can calculate the change in temperature.

Once we have the heat released by the reaction, we can use the equation q=nH to solve for H, the molar enthalpy of neutralization. This will give us the energy released per mole of sulphuric acid involved in the reaction.

I hope this helps you get started on solving the problem. Let me know if you have any further questions. Good luck!

 

[Blueshift5]

To calculate the molar enthalpy of neutralization for sulfuric acid, we first need to understand the concept of calorimetry. Calorimetry is the measurement of heat transfer in a chemical reaction. In this case, we can use the formula q=nH, where q is the heat transfer, n is the number of moles of the substance, and H is the molar enthalpy of the reaction.

To start solving this problem, we need to first determine the number of moles of sulfuric acid and sodium hydroxide that were used in the reaction. We can do this by using the formula n=CV, where C is the molar concentration of the solution and V is the volume in liters.

For sulfuric acid:
n = (1.5 mol/L) x (0.2 L) = 0.3 moles

For sodium hydroxide:
n = (1.5 mol/L) x (0.4 L) = 0.6 moles

Next, we need to calculate the heat transfer (q) in the reaction. We can use the formula q=mc(delta)T, where m is the mass of the solution, c is the specific heat capacity of the solution, and (delta)T is the change in temperature.

For sulfuric acid:
q = (0.2 kg) x (4.18 J/g*C) x (34.06-19.67) = 1217.2 J

For sodium hydroxide:
q = (0.4 kg) x (4.18 J/g*C) x (34.06-19.67) = 2434.4 J

Now, we can use the formula q=nH to solve for the molar enthalpy of neutralization (H). Since the heat transfer is the same for both solutions, we can set up an equation:

(0.3 moles)(H) = 1217.2 J
(0.6 moles)(H) = 2434.4 J

Solving for H, we get:
H = 4057.3 J/mol

Therefore, the molar enthalpy of neutralization for sulfuric acid is 4057.3 J/mol.