is that a -56? Dr. BOb. thank you
When 27.20mL of .5 M H2SO4 is added to 27.20 of 1M KOH in a coffee cup calorimeter at 23.5 degrees C, the temp rises to 30.17 degrees C. Calculate delta H of this reaction.
(Assume that the total volume is the sum of the individual volumes and that density and specific heat capacity of the solution are the same as pure water d of water =1.00g/mL: c of water =4.184J/g*c))
my work: correct or no??
• 2KOH + H2SO4→ 2H2O + K2SO4
n(KOH) = 0.0272 * 1 = 0.0272mol
n(H2SO4) =0.0272 * 0.5 = 0.0136 mol
total volume = 27.2 + 27.2 = 54.4mL
mass of solution = 54.4mL * 1g/mL =54.4g
Q = mcΔT
= 54.4g * 4.18 J/g°C * (30.17 - 23.5)°C
= 1516.74J
The heat is lost from system to surrounding, so ΔH is-ve
For every 0.0136 mol of H2SO4 react, it releases 1516.74J of heat
For 1 mole of H2SO4,
ΔH= 1516.74/0.0136
= -111.525 kJ / mol
= -112 kJ/ mol.......thank you, sir•chemistry/check my work I have to submit today - DrBob222, Saturday, October 17, 2015 at 1:06am
The heat is lost from system to surrounding, so ΔH is-ve
For every 0.0136 mol of H2SO4 react, it releases 1516.74J of heat
For 1 mole of H2SO4,
ΔH= 1516.74/0.0136
= -111.525 kJ / mol
= -112 kJ/ mol.......thank you, sIr
This value is for 2 mols H2O so you need to to do -112 kJ/mol x 1/2 = about 56 kJ/mol.
•chemistry/check my work I have to submit today - patrick, Saturday, October 17, 2015 at 2:14pm
Is that a -56 ?
To calculate the change in enthalpy (ΔH) of the reaction, we can use the formula:
ΔH = Q/n
Where Q is the heat released or absorbed by the reaction and n is the number of moles of the limiting reactant.
In this case, the limiting reactant is H2SO4, and we have calculated the number of moles of H2SO4 to be 0.0136 mol.
Q can be calculated using the formula:
Q = mcΔT
Where m is the mass of the solution, c is the specific heat capacity of water (4.184 J/g°C), and ΔT is the change in temperature (30.17°C - 23.5°C = 6.67°C).
Using the given values, we substitute them into the equation to find Q:
Q = 54.4g * 4.184 J/g°C * 6.67°C = 1516.74 J
Now we can calculate ΔH using the equation:
ΔH = 1516.74 J / 0.0136 mol = -111.525 kJ/mol
Since this value is for 2 moles of H2O, we need to divide it by 2 to get the value for 1 mole of H2O:
-111.525 kJ/mol / 2 = -55.7625 kJ/mol
Rounding to the nearest whole number, we get approximately -56 kJ/mol.
So, yes, the value for ΔH is approximately -56 kJ/mol.