Answer to Question #135505 in General Chemistry for Mary Joy Laquindanum

Question

3. For the reaction
Na2CO3 (s) + 2HCl(g) → NaCl(s) + CO2 (g) + H2O(l)
∆H is -144.1 kJ. What is ∆U.

Answer 1

We know according to thermodynamics,

∆H = ∆U +∆n.RT____[A]

Where , ∆n= difference in mole no. Of gasious reactant from the number of moles of gaseous product.

R = universal gas constant

T = absolute temperature of the system

This reaction occurs in room temperature.

i.e T=( 25 + 273) K

= 298 K

∆n =number of moles of CO2 - number of moles of HCl

= 1 - 2

= -1

Putting the given values in the above [A] equation become,

-144.1 ×10^3 = ∆U +(-1)×8.314×298

Or,∆U = -144.1×10^3 + (1×8.314×298)

=-141697.254 joul

≈-141.697 kJ

Hence, the ∆U value of the given reaction is = -141.697 kJ