When 0.100 L of 1.00 mol L−1 HCl(aq) is mixed with 0.100 L of 1.00 mol L−1 NaOH, the temperature of the solution rises by 8.5 oC. Assuming no heat loss to the container or the surroundings, what is the heat of reaction for the following reaction? (You may also assume that, for these aqueous solutions, the density is about 1.0 g mL−1 and the heat capacity is about 4.2 J g−1 oC−1.)
H+(aq) + OH−(aq) → H2O(l)
A −4.8 kJ mol−1
B −7.1 kJ mol−1
C −18 kJ mol−1
D −36 kJ mol−1
*E −71 kJ mol
I get 71 kJ/mol
q = mass H2O x specific H2O x delta T
q = 200 g x 4.2 J/g*c x 8.5 = 7140 J.
mols = M x L = 1 M x 0.1 L = 0.1
-7140 J/0.1 mol = 71400 J/mol = 71.4 kJ/mol
To calculate the heat of reaction for the given reaction, we can use the equation:
q = mcΔT
Where:
q = heat energy
m = mass
c = specific heat capacity
ΔT = change in temperature
First, let's calculate the mass of the solution. Since the density is approximately 1.0 g mL−1, the mass of 0.100 L of solution can be calculated as:
Mass = volume × density = 0.100 L × 1.0 g mL−1 × 1000 mL/L = 100 g
Next, we calculate the total heat energy absorbed or released by the solution using the equation:
q = mcΔT
The specific heat capacity, c, is given as 4.2 J g−1 oC−1.
Substituting the values into the equation, we have:
q = 100 g × 4.2 J g−1 oC−1 × 8.5 oC = 3570 J
Since 1 kJ = 1000 J, we convert the unit to kJ:
q = 3570 J ÷ 1000 = 3.57 kJ
Finally, since 1 mole of HCl reacts with 1 mole of NaOH, the heat of reaction can be calculated by dividing the heat energy by the number of moles of the limiting reactant.
In this case, both HCl and NaOH have the same concentration of 1.00 mol L−1. So, the number of moles of HCl and NaOH are:
moles = concentration × volume = 1.00 mol L−1 × 0.100 L = 0.100 mol
Therefore, the heat of reaction is:
Heat of reaction = q ÷ moles = 3.57 kJ ÷ 0.100 mol = 35.7 kJ mol−1
So, the correct answer is D. -35.7 kJ mol−1.
To determine the heat of reaction for the given reaction, we need to use the equation:
q = mcΔT
where:
q is the heat transferred (in Joules),
m is the mass of the solution (in grams),
c is the specific heat capacity of the solution (in J/g°C), and
ΔT is the change in temperature (in °C).
First, we need to determine the mass of the solution. Since the density is approximately 1.0 g/mL, we can assume that 0.100 L of the solution has a mass of 0.100 kg or 100 g.
Next, we need to calculate the change in temperature. Given that the temperature of the solution rises by 8.5°C, we have ΔT = 8.5°C.
Now, we need to calculate the heat transferred (q). Using the equation mentioned above, we have:
q = (100 g) * (4.2 J/g°C) * (8.5°C)
q = 3570 J
Since the reaction is between 1 mol of HCl and 1 mol of NaOH, the molar heat of reaction can be calculated by dividing the heat transferred by the number of moles reacting. In this case, 1 mol of HCl and 1 mol of NaOH reacts, so the molar heat of reaction is:
Molar heat of reaction = (3570 J) / (1 mol + 1 mol)
Molar heat of reaction = 3570 J/mol
To convert the molar heat of reaction to kJ/mol, we divide by 1000:
Molar heat of reaction = 3570 J/mol / 1000
Molar heat of reaction = 3.57 kJ/mol
Since the molar heat of reaction is the negative of the enthalpy change, the correct answer is:
E) The heat of reaction for the given reaction is approximately -3.57 kJ/mol or -71 kJ mol^-1.