(1) MgCO3(s) ⟶ MgO(s) + CO2(g) ΔrH = 118.1 kJ mol−1
(2) Mg(s) + 12O2(g) ⟶ MgO(s) ΔfH =−601.7 kJ mol−1
What is the enthalpy change, ΔrH for the following reaction?
2 MgCO3 ⟶ 2 Mg(s) + 2 CO2(g) + O2(g)
I didn't write this down but tried to do it in my head; therefore, I may have made a mistake so check this out thoroughly.
Multiply eqn 1 by 2 and add it to the reverse of equn 2 multiplied by 2. See if that will give you the final equation you want. Change the sign of dH when reversing.
so, 118x-601 + 2x601?
no i dont think thats right
You're right, it isn't right. For two reasons.
First your second equation is not balanced.
Since it isn't balanced those instructions I gave can't be right.
Also you didn't follow instructions and check the final equation. That equation was not correct.
Multiply eqn 1 by 2 and add to the reverse of eqn 2.
Here is 1 and corrected 2.
(1) MgCO3(s) ⟶ MgO(s) + CO2(g) ΔrH = 118.1 kJ mol−1
(2) 2Mg(s) + O2(g) ⟶ 2MgO(s) ΔfH =−601.7 kJ mol−1 and you want
2 MgCO3 ⟶ 2 Mg(s) + 2 CO2(g) + O2(g)
Follow the instructions.
eqn1 x 2. 2MgCO3(s) ⟶ 2MgO(s) + 2CO2(g) ΔrH =118.1 kJ mol−1 x 2 =?
rev eqn 2. 2MgO(s) ==> 2Mg(s) + O2(g) −601.7 kJ mol−1 x -1 = ?
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add to get 2 MgCO3 ⟶ 2 Mg(s) + 2 CO2(g) + O2(g) (118.1*2) + 601,7 = ?
To find the enthalpy change, ΔrH, for the reaction:
2 MgCO3 ⟶ 2 Mg(s) + 2 CO2(g) + O2(g)
We can use the given enthalpy changes for the reactions (1) and (2) and apply Hess's law. Hess's law states that if a reaction can be expressed as the sum of two or more reactions, the enthalpy change for the overall reaction is equal to the sum of the enthalpy changes for each individual reaction.
Step 1: Rearrange the given reactions so that they match the desired reaction:
MgO(s) + CO2(g) ⟶ MgCO3(s) (1)
Mg(s) + 1/2 O2(g) ⟶ MgO(s) (2)
Step 2: Multiply reaction (1) by 2 to match the number of moles of MgCO3 in the desired reaction:
2 MgO(s) + 2 CO2(g) ⟶ 2 MgCO3(s) (3)
Step 3: Multiply reaction (2) by 2 to match the number of moles of Mg in the desired reaction:
2 Mg(s) + O2(g) ⟶ 2 MgO(s) (4)
Step 4: Add reactions (3) and (4):
2 MgO(s) + 2 CO2(g) + 2 Mg(s) + O2(g) ⟶ 2 MgCO3(s) + 2 MgO(s)
Step 5: Cancel out the common species on both sides of the equation:
2 MgO(s) + 2 CO2(g) + 2 Mg(s) + O2(g) ⟶ 2 MgCO3(s) + 2 MgO(s)
(canceling MgO and MgCO3)
2 CO2(g) + O2(g) ⟶ 2 Mg(s)
Now, we have the desired reaction:
2 MgCO3 ⟶ 2 Mg(s) + 2 CO2(g) + O2(g)
Step 6: Calculate the overall enthalpy change by summing the individual enthalpy changes:
ΔrH = ΔH (reaction (3)) + ΔH (reaction (4))
ΔrH = 2 * (ΔH (reaction (1))) + 2 * (ΔH (reaction (2)))
ΔrH = 2 * (118.1 kJ mol^−1) + 2 * (-601.7 kJ mol^−1)
ΔrH = 236.2 kJ mol^−1 + (-1203.4 kJ mol^−1)
ΔrH = -967.2 kJ mol^−1
Therefore, the enthalpy change, ΔrH, for the given reaction is -967.2 kJ mol^−1.