ThermochemistryHard
Question
Calculate the enthalpy of formation (in kcal/mol) of gaseous HCl using the following data.
Substance NH3(g) HCl(g) NH4Cl(s)
Heat of formation −11 X −75
Heat of solution −8.5 −17.5 +3.9 kcal
NH3 (aq) + HCl (aq) → NH4Cl (aq): ΔH = −12 kcal
Options
A.−44.2
B.−22.1
C.−11.05
D.−28.7
Solution
$- 12 = \left\lbrack ( - 75) + (3.9) \right\rbrack - \left\lbrack ( - 11) + ( - 8.5) + x + ( - 17.5) \right\rbrack$
$\therefore X = - 22.1\text{ kcal/mol}$
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