Q1) How much heat is released when 6.38 grams of Ag(s) (m. m = 107.9 g/mol) reacts by the equation shown below at standard state conditions?
4A9 (s) + 2H, Sq) + O2(g)
2Ag $(s) + 2H200)
Substance
AHof (kJ/mol)
-20.6
H259)
Ag2S (5)
H200
-32.6
-285.8
a)
8.80 KI
b) 69.9 kJ
C) 22.1 kJ
d) 90.8 kJ
e) 40.5 kJ

The correct answer is -8.80 kJ.

Explanation:

The ΔH° can be determined by using the formula,  

ΔH°rxn = ΔH°f (products) - ΔH°f(reactants)

Based on the given information, the ΔH°f of H2S(g) is -20.6, for Ag2S (s) is -32.6 and for H2O (l) is -285.8 kJ/mole.  

Now putting the values we get,  

= [2 molΔH°f (Ag2S) + 2 molΔH°f (H2O)] - [4 molΔH°f(Ag) + 2 molΔH°f(H2S) + 1 molΔH°f(O2)]

=[2 mol (-32.6 kJ/mol) + 2 mol(-285.8 kJ/mol)] - [4 mol(0.00 kJ/mol) + 2 mol (-20.6 kJ/mol) + 1 mol (0.00 kJ/mol)

= [(-65.2 kJ) + (-571.6 kJ)] - [(-41.2 kJ)]

= -595.6 kJ

Thus, the enthalpy change of -595.6 kJ takes place when 4 mol of Ag reacts by the equation mentioned.  

The mass of Ag given is 6.38 grams, the molecular mass of Ag is 107.9 g/mol. The formula for calculating moles is,  

Moles = mass/molar mass

= 6.38 g / 107.9 g/mol

= 0.0591 mol

Now the change in enthalpy when 0.0591 mol of Ag reacts by the given reaction is (-595.6 kJ/4 mol) × 0.0591 mol = -8.80 kJ

The negative sign indicates that the heat is released in the process. Therefore, the -8.80 kJ of heat is released by 6.38 grams of Ag in the given case.  

vanderwaal's forces of attraction are the electrostatic intermolecular forces that exist in a molecule due to the temporary or permanent dipoles that are developed in a molecule. in , the major intermolecular forces are dipole-dipole interactions as the molecule is polar (bent structure). so the vanderwaal's forces of interaction in between molecules are the permanent dipole-dipole interactions that develop due to the dipole that exists in between the polar h - s bond. the positively charged hydrogen atom is attracted towards the negative end of the dipole that is sulfur of the adjacent molecule. therefore, these forces of attraction are relatively stronger than that of london dispersion forces that exist in non polar molecules.