Calculate ΔHrxn for the following reaction:
CaO(s)+CO2(g)→CaCO3(s)
Use the following reactions and given ΔH′s.
Ca(s)+CO2(g)+12O2(g)→CaCO3(s), ΔH= -812.8 kJ
2Ca(s)+O2(g)→2CaO(s), ΔH= -1269.8 kJ
dHrxn = (n*dHf products( - (n*dHf reactants)
Substitute and solve
To calculate ΔHrxn for the given reaction, we can use the concept of Hess's Law. According to Hess's Law, the enthalpy change of a reaction is the same regardless of the pathway taken.
In this case, we need to manipulate the given equations to obtain the desired reaction equation. We can do so by combining the given equations and reversing them if necessary. Here's how we can proceed:
1. Reverse the second equation:
2CaO(s) → 2Ca(s) + O2(g)
ΔH = +1269.8 kJ (since we have reversed the equation, sign changes)
2. Multiply the first equation by 2 to balance the number of CaO(s) on both sides and reverse it:
2CaCO3(s) → 2Ca(s) + 2CO2(g) + 12O2(g)
ΔH = -2 × (-812.8 kJ) = +1625.6 kJ (since we have reversed the equation, sign changes)
3. Finally, add the two modified equations together:
2Ca(s) + O2(g) + 2CaCO3(s) → 2Ca(s) + 2CO2(g) + 12O2(g)
ΔHrxn = ΔH1 + ΔH2
= (+1269.8 kJ) + (+1625.6 kJ)
= +2895.4 kJ
Therefore, the ΔHrxn for the given reaction is +2895.4 kJ.
To calculate ΔHrxn for the reaction CaO(s)+CO2(g)→CaCO3(s), we can use the given reactions and their respective ΔH values as follows:
1. Multiply the first given reaction by 2 to match the coefficients of CaO in the desired reaction:
2[Ca(s) + CO2(g) + 1/2O2(g) → CaCO3(s)] with ΔH = -2 * 812.8 kJ
2. Multiply the second given reaction by 2 and reverse it to match the coefficients of CaO in the desired reaction:
2CaO(s) → 2Ca(s) + O2(g) with ΔH = -1 * (-1269.8) kJ
3. Add the modified equations together:
2[Ca(s) + CO2(g) + 1/2O2(g) → CaCO3(s)] + 2CaO(s) → 2Ca(s) + O2(g)
(2 * -812.8 kJ) + (2 * -(-1269.8) kJ)
= -1625.6 kJ + 2539.6 kJ
= 914 kJ
Therefore, the ΔHrxn for the reaction CaO(s) + CO2(g) → CaCO3(s) is 914 kJ.