The boiling point and melting point of substances is determined by their molecular weight and the cohesion (stickiness) of the individual molecules in the substance.
Based on the cohesive molecule - molecule interactions of a substance, it may have either a high melting and boiling point or vice versa. The stronger the cohesive interactions the more the heat required for the substance's transition into liquid and gas state.
These intermolecular forces include hydrogen bonding , ionic bonding, dipole dipole, and Van Der Waals dispersion forces.
Answer:
The enthalpy change during the reaction is -199. kJ/mol.
Explanation:
![Mn(s)+2HCl(aq)\rightarrow MnCl_2(aq)+H_2(g)](https://tex.z-dn.net/?f=Mn%28s%29%2B2HCl%28aq%29%5Crightarrow%20%20MnCl_2%28aq%29%2BH_2%28g%29)
Mass of solution = m
Volume of solution = 100.0 mL
Density of solution = d = 1.00 g/mL
![m=1.00 g/mL\times 100.0 mL = 100 g](https://tex.z-dn.net/?f=m%3D1.00%20g%2FmL%5Ctimes%20100.0%20mL%20%3D%20100%20g)
First we have to calculate the heat gained by the solution in coffee-cup calorimeter.
![q=m\times c\times (T_{final}-T_{initial})](https://tex.z-dn.net/?f=q%3Dm%5Ctimes%20c%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29)
where,
m = mass of solution = 100 g
q = heat gained = ?
c = specific heat = ![4.18 J/^oC](https://tex.z-dn.net/?f=4.18%20J%2F%5EoC)
= final temperature = ![23.1^oC](https://tex.z-dn.net/?f=23.1%5EoC)
= initial temperature = ![28.9^oC](https://tex.z-dn.net/?f=28.9%5EoC)
Now put all the given values in the above formula, we get:
![q=100 g \times 4.18 J/^oC\times (28.9-23.1)^oC](https://tex.z-dn.net/?f=q%3D100%20g%20%5Ctimes%204.18%20J%2F%5EoC%5Ctimes%20%2828.9-23.1%29%5EoC)
![q=2,242.4 J=2.242 kJ](https://tex.z-dn.net/?f=q%3D2%2C242.4%20J%3D2.242%20kJ%20)
Now we have to calculate the enthalpy change during the reaction.
![\Delta H=-\frac{q}{n}](https://tex.z-dn.net/?f=%5CDelta%20H%3D-%5Cfrac%7Bq%7D%7Bn%7D)
where,
= enthalpy change = ?
q = heat gained = 2.242 kJ
n = number of moles fructose = ![\frac{\text{Mass of manganese}}{\text{Molar mass of manganese}}=\frac{0.620 g}{54.94 g/mol}=0.0113 mol](https://tex.z-dn.net/?f=%5Cfrac%7B%5Ctext%7BMass%20of%20manganese%7D%7D%7B%5Ctext%7BMolar%20mass%20of%20manganese%7D%7D%3D%5Cfrac%7B0.620%20g%7D%7B54.94%20g%2Fmol%7D%3D0.0113%20mol)
![\Delta H=-\frac{2.242 kJ}{0.0113 mol }=-199. kJ/mol](https://tex.z-dn.net/?f=%5CDelta%20H%3D-%5Cfrac%7B2.242%20kJ%7D%7B0.0113%20mol%20%7D%3D-199.%20kJ%2Fmol)
Therefore, the enthalpy change during the reaction is -199. kJ/mol.