when 50.0 ml of 2.00M KOH is reached with 50.0 ml of 2.00M HCl, the temperature rose from 20.0 degree celcus to 28.7 degree celcus. What is the heat generated by this neutralization? (Assume that the density fo the two 2 M solution is 2.00g/ml and that the heat capacity of the reaction solution si 4.184 j/degree celcus g.)
Please provide steps to get to answer
heat= total mass*heatcapacity*deltaTEmp
= total volume*density*heatcapacity*deltaTemp
8.3m KBr with a density of 1.22g/mL
To find the heat generated by this neutralization reaction, we need to use the formula:
Q = m × C × ΔT
Where:
Q is the heat generated (in Joules),
m is the mass of the solution (in grams),
C is the heat capacity of the solution (in Joules per degree Celsius per gram), and
ΔT is the change in temperature (in degrees Celsius).
Let's calculate step by step:
Step 1: Calculate the mass of the solution
Since we are given the volumes of the KOH and HCl solutions, we can calculate their masses using their densities.
Mass of KOH solution = volume × density = 50.0 ml × 2.00 g/ml = 100.0 g
Mass of HCl solution = volume × density = 50.0 ml × 2.00 g/ml = 100.0 g
The total mass of the solution = mass of KOH solution + mass of HCl solution = 100.0 g + 100.0 g = 200.0 g
Step 2: Calculate the change in temperature (ΔT)
ΔT = final temperature - initial temperature = 28.7 °C - 20.0 °C = 8.7 °C
Step 3: Substitute the values into the formula
Q = m × C × ΔT
= 200.0 g × 4.184 J/°C g × 8.7 °C
Step 4: Calculate the heat generated
Q ≈ 7308 J
Therefore, the heat generated by this neutralization reaction is approximately 7308 Joules.