please assist me with this themochemistry question
Given the following data
CaCO3 (s)>>>>>>CaO (s) + CO2(g) (g) Delta H = -178.1 kj
C (s) + O2>>>>>>CO2 (g) Delta H = -393.5 kj
Calculate enthalpy in kj of the following reaction
CaCO3 (s)>>>>>>>>>CaO (s) + C (s) + O2 (g)
a) 215.4
b) 571.6
c) -215.4
d) -571.6
e) 178.
Add equation 1 to the reverse of equation 2 and you will obtain the reaction you want. Change the sign of delta H for equation 2 (since it is reversed) and add it to delta H for equation 1 to obtain delta H for the desired reaction.
htryh
To calculate the enthalpy change of the given reaction, we need to use the Hess's Law. Hess's Law states that the overall enthalpy change of a reaction is equal to the sum of the enthalpy changes of the individual steps.
Step 1: CaCO3 (s) >>>>>> CaO (s) + CO2 (g) (g) ΔH = -178.1 kJ
Step 2: C (s) + O2 >>>>>> CO2 (g) ΔH = -393.5 kJ
The target reaction is the sum of Step 1 and Step 2 after adjusting the coefficients to match the reaction:
CaCO3 (s) >>>>>> CaO (s) + C (s) + O2 (g)
To calculate the enthalpy change of the target reaction, we add the enthalpy changes of Step 1 and Step 2.
ΔH(target) = ΔH(Step 1) + ΔH(Step 2)
ΔH(target) = -178.1 kJ + (-393.5 kJ)
ΔH(target) = -571.6 kJ
Therefore, the enthalpy change of the given reaction is -571.6 kJ.
The correct answer is option d) -571.6.