17. Calculate ƒ´H of strontium carbonate using Hess law
Sr + C (graphite) + 3/2 O2 ¡÷ SrCO3
Experimental information
SrO ¡÷ Sr + 1/2 O2 ƒ´H 592 kJ
SrO + CO2 ¡÷ SrCO3 ƒ´H -234 kJ
2 C (graphite) + 2 O2 ¡÷ 2 CO2 ƒ´H -788 kJ
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I've shown you how to do these above. Try your hand at this. There is little I can do except do it for you and I don't want to do that. If you still have trouble explain in detail what your problem is and I'll try to help you through it.
To calculate ΔH of strontium carbonate (SrCO3) using Hess's Law, we need to determine the overall ΔH by combining the known reactions in the appropriate way.
The given reactions are:
1) SrO ⟶ Sr + 1/2 O2 ΔH = 592 kJ
2) SrO + CO2 ⟶ SrCO3 ΔH = -234 kJ
3) 2 C (graphite) + 2 O2 ⟶ 2 CO2 ΔH = -788 kJ
To calculate the ΔH of the target reaction, which is Sr + C(graphite) + 3/2 O2 ⟶ SrCO3, we need to manipulate and combine the given reactions in such a way that the target reaction is produced.
First, let's reverse reaction 1) to get Sr + 1/2 O2 ⟶ SrO and then multiply it by -1 to change the sign of ΔH.
-1(SrO ⟶ Sr + 1/2 O2) ΔH = -592 kJ
Next, let's multiply reaction 2) by 1/2 to get 1/2 SrO + 1/2 CO2 ⟶ 1/2 SrCO3 and change the sign of ΔH.
(1/2)(SrO + CO2 ⟶ SrCO3) ΔH = 1/2(-234 kJ)
Finally, let's multiply reaction 3) by 3 to get 6 C (graphite) + 6 O2 ⟶ 6 CO2 and change the sign of ΔH.
3(2 C (graphite) + 2 O2 ⟶ 2 CO2) ΔH = 3(-788 kJ)
Now, we can add these equations together to get the target reaction:
Sr + 1/2 O2 + 1/2 SrO + 1/2 CO2 + 6 C (graphite) + 6 O2 ⟶ SrCO3 + 6 CO2
The overall ΔH of the target reaction is the sum of the ΔH values from each equation.
ΔH = (-592 kJ) + [1/2(-234 kJ)] + [3(-788 kJ)]
Now, calculate the numerical value of ΔH using the given ΔH values:
ΔH = -592 kJ + (-117 kJ) + (-2364 kJ)
Simplifying:
ΔH = -3073 kJ
Therefore, the ΔH of the reaction Sr + C(graphite) + 3/2 O2 ⟶ SrCO3 is -3073 kJ.