A total of 2.00 mol of a compound is allowed to react in a foam coffee cup that contains 141 g of water. The reaction caused the temperature of the water to rise from 21.0 to 24.7 degrees Celsius. What is the enthalpy of this reaction? Assume that no heat is lost to the surroundings or to the coffee cup itself?
q = mass x specific heat x delta T.
That gives you q (delta H) for 2.00 mol. Half that will be the delta H/1 mol.
so what's the answer?
what's the answer?
To find the enthalpy of the reaction, we can use the formula:
q = m * c * ΔT
where:
q is the heat absorbed or released by the water (in Joules or J)
m is the mass of the water (in grams or g)
c is the specific heat capacity of water (4.18 J/g°C)
ΔT is the change in temperature of the water (in degrees Celsius or °C)
First, we need to calculate the heat absorbed or released by the water using the given information.
The mass of the water is given as 141 g, the specific heat capacity of water is 4.18 J/g°C, and the change in temperature is from 21.0 to 24.7 degrees Celsius:
ΔT = 24.7°C - 21.0°C
ΔT = 3.7°C
q = 141 g * 4.18 J/g°C * 3.7°C
q = 2109.6 J
Now, we need to convert the moles of the compound to the number of moles reacted per mole of water.
Since the compound is given in moles and assuming it reacts in a 1:1 ratio with water:
moles of water = 2.00 mol
Next, we need to calculate the molar enthalpy change by dividing the heat absorbed or released by the number of moles of water:
ΔH = q / moles of water
ΔH = 2109.6 J / 2.00 mol
ΔH = 1054.8 J/mol
Therefore, the enthalpy of this reaction is 1054.8 J/mol.