I realize some of these you cannot do but do the best you can with the informati
ID: 801065 • Letter: I
Question
I realize some of these you cannot do but do the best you can with the information provided thanks!
Experimental Produce, Part B. Three student chemists measured 50.0 mL of 1.00 M NaOH in separate Styrofoam coffee cup calorimeters (Part B). Brett added 50.0 mLof 1.10 M HCl to his solution of NaOH; dale added 45.5 mL of 1.10 m Hcl (equal moles) to his NaOH solution. Lyndsay added 50.0 mL of 1.00 M HCl to her NaOH solution. Each student recorded the temperature change and calculated the enthalpy of neutralization. Identify the student who student who observes a temperature change that will be different from that observed by the other two chemists. Explain why and how (higher or lower) the temperature will be different. Experimental Produce, Part C. Angelina observes a temperature increases when her salt dissolves in water. Is the lattice for the salt greater or less than the hydration energy for the salt? Explain. Will the solubility of the salt increases or decreases with temperature increases? Explain. A 5.00-g sample of KBr at 25.0 degree C dissolves in 25.0 mL of water also at 25.0 degree C. The final equilibrium temperature of the resulting solution is 18.1 degree C. What is the enthalpy of solution, Delta H, of KBr expressed in kilojoules per mole?Explanation / Answer
4) Dale is adding the stoichiometric amount of acid in a smaller volume of water, so there is less water to heat up, so the temperature rise will be (a little) more for Dale.
5) a)The water is supplying the energy to break the lattice bonds (which would cool down the water). Surrounding ions with water gives energy, which heats up the water (Can also think of it as increasing entropy). Since the salt dissolves and the water heats up, the hydration energy must be the larger of the two.
b) La Chatelier's Principle can be used here. Heat is a product, so if we increase heat, the equilibrium will shift back to reactants. This means decreased solubility. (but rate getting to the lower equilibrium concentration will speed up)
6) 5.00 g is 5.00/(39.1+79.9) = 0.042 moles. Energy increase is 25g*(1 cal/g deg)(18.1 - 25)deg = -172.5 cal = -721.7 J
this is -721.7J/0.042 moles = -17,177 J/mol = -17.2 kj/mol