Question

Limestone stalactites and stalagmites are formed in caves by the following reaction: Ca2 (aq) 2HCO−3(aq)→CaCO3(s) CO2(g) H2O(l) If 1 mol of CaCO3 forms at 298 K under 1 atm pressure, the reaction performs 2.47 kJ of P−V work, pushing back the atmosphere as the gaseous CO2 forms. At the same time, 38.65 kJ of heat is absorbed from the environment. What is the value of ΔE for this reaction

Answer 1

Answer : The value of $\Delta E$ for this reaction is 36.18 kJ

Explanation :

First law of thermodynamic : It states that the energy can not be created or destroyed, it can only change or transfer from one state to another state.

As per first law of thermodynamic,

$\Delta E=q+w$

where,

$\Delta E$ = internal energy  of the system

q = heat added or rejected by the system

w = work done

As we are given that:

q = 38.65 kJ

w = -2.47 kJ (system work done on surrounding)

Now put all the given values in the above expression, we get:

$\Delta E=38.65kJ+(-2.47kJ)$

$\Delta E=36.18kJ$

Therefore, the value of $\Delta E$ for this reaction is 36.18 kJ