In a coffee cup calorimeter, 50.00 mL of HNO3 (aq) and 50.00 mL of KOH(aq) were mixed. The temperature of the resultant solution increased 6.5 C. What is the Delta H of this reaction?

heat = masstotal*specificheatwater*6.5

= .050kg*2*6.5C*specific heat

so look up the specific heat of water in units of joules/kg-C

specific heat of water 4.185? Look in the back of your chemistry book or use the index to find an example

To determine the change in enthalpy (ΔH) of a reaction using a coffee cup calorimeter, we can apply the principle of heat transfer, which states that the heat gained by the solution is equal to the heat lost by the surroundings. In this case, the surroundings refer to the coffee cup calorimeter.

To calculate the heat gained or lost by the solution, we need to use the formula:

q = m * c * ΔT

Where:
q = heat gained or lost by the solution
m = mass of the solution (in grams)
c = specific heat capacity of the solution (in J/g°C)
ΔT = change in temperature of the solution (in °C)

Since the volumes of the nitric acid (HNO3) and potassium hydroxide (KOH) are given, we assume their densities to be equal to 1 g/mL, which means 50.00 mL of each solution equals 50.00 g.

The specific heat capacity of water is commonly used as an approximation for dilute aqueous solutions, so we can assume the specific heat capacity (c) of the solution is 4.18 J/g°C.

Now, we can plug in the values:

q = (mHNO3 + mKOH) * c * ΔT

q = (50.00 g + 50.00 g) * 4.18 J/g°C * 6.5°C

Simplifying:

q = 100.00 g * 4.18 J/g°C * 6.5°C

q = 2707 J

Hence, the heat gained or lost by the solution is 2707 J.

Since the heat gained by the solution is equal to the heat lost by the surroundings (the coffee cup calorimeter), we can say:

q solution = -q calorimeter

Therefore, the ΔH of the reaction (q reaction) can be calculated as -2707 J.

So, the ΔH of the reaction when mixing 50.00 mL of HNO3 (aq) and 50.00 mL of KOH(aq) in the coffee cup calorimeter is -2707 J.