Consider this equation. 2CaCO 3 (s) --> 2CaO (s) + 2 CO 2 (g) ?H = -178.1 kJ /mo
ID: 1036697 • Letter: C
Question
Consider this equation.
2CaCO3(s) --> 2CaO(s) + 2 CO2(g) ?H = -178.1 kJ /molrxn
If 2.50 x 102 grams CaCO3 is reacted at a constant the pressure of 0.800 atm and temperature of 200 K, what was the change in internal energy (?E) in this process, assuming the reaction goes to completion? Assume solids have negligible volume.
- 8.05 x 102 kJ
37.8 kJ
- 5.84 kJ
- 17.2 kJ
-232 kJ
- 503 kJ
296.0 kJ
- 3.19 x 102 kJ
- 8.05 x 102 kJ
37.8 kJ
- 5.84 kJ
- 17.2 kJ
-232 kJ
- 503 kJ
296.0 kJ
- 3.19 x 102 kJ
Explanation / Answer
2CaCO3(s) --> 2CaO(s) + 2 CO2(g) ; given del(H) = -178.1kj/molrxn
moles of CaCO3 = 2.5*102/100 = 2.5 ; del(H) = -178.1*0.025*102*2 = -8. 9*102kj
H = U+PV ; del(H) = del(U)+RTdel(n) = del(U)+8.314*2*473 = -8.05*102kj ; no change in V and P
So del(U) = -8. 05*102kj