Question

Na2CO3 (s)  +  2HCl(g)   →  NaCl(s)  +  CO2 (g)  +  H2O(l)

      ∆H is -144.1 KJ. What is ∆E.

Answer 1

$Na_2CO_{3(s)} + 2HCl_{(g)} \to NaCl_{(s)} + CO_{2(g)} + H_2O_{(l)} \qquad ∆H = -144.1 kJ$

$∆E = ∆H - RT∆n$

$∆n = n_p - n_r = 1 - 2 = -1$

Assuming the reaction takes place at STP,

$RT∆n = 8.314×273×-1 = -2269.7J$

from the chemical reaction above, $∆H = -144.1kJ = -144,100J$

$∆E = -144100J - -(2269.7)J = -141830.3J$

$\therefore$ ∆E = -141.8kJ