Question

When 1 mol of CS₂(l) forms from its elements at 1 atm and 25°C, 89.7 kJ is absorbed, and it takes 27.7 kJ to vaporize 1 mol of the liquid. How much heat is absorbed when 1 mol of CS₂(g) forms from its elements at these conditions? Do not answer in scientific notation.

Answer 1

Solution:

Use Hess's Law to find the answer:

$C_{graphite}(s)\;+\;2S(s)\rightarrow\;CS_2(l)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\Delta H=89.7\;(\frac{kJ}{mole})$

$CS_2(l)\rightarrow CS_2(g) \;\;\;\;\;\;\;\;\;\;\;\;\;\; \Delta H=27.7(\frac{kJ}{mole})$

$C_{graphite}(s)+2S(s) \rightarrow CS_2 (g) \;\;\;\;\;\;\;\;\;\;\;\;\;\; \Delta H=89.7+27.7=117.4 (\frac{kJ}{mole})$