2CO(g) = 2C(s) + O2(g).
A sample of 0.882g of carbon monoxide, CO(g), is completely decomposed in a calorimeter containing 325 g of water. The initial temperature is 17.60*C, and the final temperature is 74.21*C. Calculate the enthalpy change in J/mol of reactant.
I hope that's balanced correctly..
q = mass H2O x specific heat H2O x (Tfinal-=Tinitial)
You know mass H2O, you can look up specific heat H2O and you are given Tfinal and Tinitial. Substitute and solve for q and that will be enthalpy for 0.882 g CO or J/0.882. Convert that to J/mol.
J/0.882 x (28g/mol) = dH in J/mol.
Yes, the equation you provided is correctly balanced. To calculate the enthalpy change in J/mol of reactant, we need to use the formula:
ΔH = q/n
where:
ΔH = enthalpy change in J/mol
q = heat absorbed or released during the reaction in J
n = number of moles of reactant
To find the heat absorbed or released during the reaction, we can use the formula:
q = mcΔT
where:
q = heat absorbed or released in J
m = mass of water in g
c = specific heat capacity of water (4.184 J/g°C)
ΔT = change in temperature in °C
First, we need to calculate the heat absorbed or released by the water:
q = mcΔT
= (325 g)(4.184 J/g°C)(74.21°C - 17.60°C)
= 90,176.92 J
Next, we need to calculate the number of moles of CO(g) using the molar mass of CO:
Molar mass of CO = 12.01 g/mol + 16.00 g/mol = 28.01 g/mol
Number of moles (n) of CO = mass/molar mass
= 0.882 g/28.01 g/mol
= 0.0315 mol
Finally, we can calculate the enthalpy change:
ΔH = q/n
= 90,176.92 J/0.0315 mol
= 2,857,222.86 J/mol
Therefore, the enthalpy change in J/mol of reactant (CO(g)) is approximately 2,857,222.86 J/mol.
To calculate the enthalpy change in J/mol of the reactant, we need to use the equation:
ΔH = q / n
where:
ΔH is the enthalpy change (in J/mol)
q is the heat absorbed by the water (in J)
n is the number of moles of the reactant
First, we need to find the number of moles of carbon monoxide, CO(g), using its molar mass.
The molar mass of CO is the sum of the molar masses of carbon (C) and oxygen (O):
Molar mass of C = 12.01 g/mol
Molar mass of O = 16.00 g/mol
Molar mass of CO = Molar mass of C + Molar mass of O = 12.01 g/mol + 16.00 g/mol = 28.01 g/mol
Now, let's calculate the number of moles of CO:
Number of moles = Mass / Molar mass = 0.882 g / 28.01 g/mol = 0.0315 mol
Now, to find the heat absorbed by the water, we can use the formula:
q = mcΔT
where:
q is the heat absorbed by the water (in J)
m is the mass of water (in g)
c is the specific heat capacity of water (4.18 J/g°C)
ΔT is the change in temperature (final temperature - initial temperature)
Substituting the values:
m = 325 g (provided in the problem)
c = 4.18 J/g°C
ΔT = 74.21°C - 17.60°C = 56.61°C
Now, let's calculate the heat absorbed by the water:
q = (325 g) × (4.18 J/g°C) × (56.61°C) = 73752.105 J
Finally, we can calculate the enthalpy change:
ΔH = q / n = 73752.105 J / 0.0315 mol = 2338333.333 J/mol
The enthalpy change for the reaction is approximately 2.34 × 10^6 J/mol.