Question

Limestone stalactites and stalagmites are formed in caves by the following reaction: $$ \mathrm{Ca}^{2+}(a q)+2 \mathrm{HCO}_{3}^{-}(a q) \longrightarrow \mathrm{CaCO}_{3}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) $$ If \(1 \mathrm{~mol}\) of \(\mathrm{CaCO}_{3}\) forms at \(298 \mathrm{~K}\) under 1 atm pressure, the reaction performs \(2.47 \mathrm{~kJ}\) of \(P-V\) work, pushing back the atmosphere as the gaseous \(\mathrm{CO}_{2}\) forms. At the same time, \(38.95 \mathrm{~kJ}\) of heat is absorbed from the environment. What are the values of \(\Delta H\) and of \(\Delta E\) for this reaction?

Short answer

For the given reaction, the internal energy change (∆E) is \(41.42 \: \text{kJ/mol}\) and the enthalpy change (∆H) is \(43.89 \: \text{kJ/mol}\).

Step-by-step solution

Recall the first law of thermodynamics formula

The first law of thermodynamics states: \[ ∆E = q + w \] where ∆E is the change in internal energy, q is the heat, and w is the work done by (or on) the system.

Calculate the internal energy change from the heat and work values

Given the heat (q) = 38.95 kJ and P-V work (w) = 2.47 kJ, we can find ∆E as follows: \[ ∆E = q + w = 38.95 + 2.47 = 41.42 \: \text{kJ/mol}\]

Recall the relationship between enthalpy change, internal energy change, and P-V work

The relationship between enthalpy change (∆H), internal energy change (∆E), and P-V work (w) is given by: \[ ∆H = ∆E + w \]

Calculate the enthalpy change using the relationships derived in Step 3

Using the calculated internal energy change (∆E = 41.42 kJ/mol) and the given P-V work (w = 2.47 kJ/mol), we can find the enthalpy change as follows: \[ ∆H = ∆E + w = 41.42 + 2.47 = 43.89 \: \text{kJ/mol}\]

Present the final answer for the enthalpy change and internal energy change in the reaction

For the given reaction, we have calculated the values of the internal energy change and the enthalpy change as follows: - The internal energy change (∆E) = 41.42 kJ/mol - The enthalpy change (∆H) = 43.89 kJ/mol