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
$- 12 = \left\lbrack ( - 75) + (3.9) \right\rbrack - \left\lbrack ( - 11) + ( - 8.5) + x + ( - 17.5) \right\rbrack$
$\therefore X = - 22.1\text{ kcal/mol}$
Create a free account to view solution
View Solution FreeMore Thermochemistry Questions
The difference between enthalpies of reaction at constant pressure and constant volume for the reaction2C6H6(l) + 15O2(g...The enthalpies of formation of SO2(g), H2O(l), HCl(g) and H2SO4(l) are −70.97, −68.32, −22.1 and −188.84 (kcal mol−1). T...The enthalpies of combustion of formaldehyde and paraformaldehyde (a polymer of formaldehyde) are −134 and −732 kcal/mol...The enthalpy of combustion of propane (C3H8) gas in terms of given data is : Bond energy (kJ/mol) εC-H εO=O &#...The enthalpy of formation of methane(g) at constant pressure is −18,500 cal/mol at 27°C. The enthalpy of formation at co...