help please Calculate the heats of dissolution for lithium chloride and ammonium

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help please

Calculate the heats of dissolution for lithium chloride and ammonium chloride from Tabulated Delta Hf values, these can be found in Appendix B of your textbook or on-line. When a 3.25-g sample of solid sodium hydroxide was dissolved in a calorimeter in 100.0 g of water, the temperature rose from 23.9 degree C to 32.0 degree C. Calculate Delta ? (in kJ/mol NaOH) for the solution process: NaOH(s) rightarrow Na+(aq) + OH-(aq) Assume it's a perfect calorimeter and that the specific heat of the solution is the same as that of pure water. Write the molecular, ionic, and net ionic equations for the following neutralization reactions: HCl(aq) + NaOH (aq) rightarrow HNO3(aq) + NaOH (aq) rightarrow Calculate the Enthalpy of neutralization for the above net ionic equation. Tabulated Delta ?f values can be found in the appendix of your textbook. Describe two methods to determine the mass of your solution in the calorimeter?

Explanation / Answer

1.when we can't find just what we want, we develop our own

the heat of solution for Lithium Chloride covers this:

LiCl (solid) --> Li+ (aq) & Cl- (aq)

all three species are usually listed in most books, though most books vary slightly in their values

LiCl (solid).... dHf = -408.3 kJ/mole

Li+ (aq) ........ dHf = -278.5 kJ

Cl- (aq) ...... dHf = -167.2

dH reaction = dHf products - dHf reactants

dH = (-278.5 & - 167.2) - ( -408.3)

dH = (-445.7) + 408.3

dH = - 37.4 kJ/mol

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2.

First, the actual amount of kJ of heat produced should be calculated, since we are only given the temperature cheg change . We can do this by finding the limiting reagent of the reaction, which is NaOH in this case.

3.25 gms ==> 3.24 /40 = 0.08125 mol NaOH

Qw= m Cp dT

= 100 * 4.2 * 8.1

=3.402 KJ/KgK

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3.sol :-

As a note, the symbols after the parentheses indicate the physical state of the substance we are regardiing.

aq = aqeous solution

l = liquid

1. The molecular equation is as follows:

HCl (aq) + NaOH (aq) --> H2O (l) + NaCl (aq)

2. The complete ionic equation (the one you're looking for) consists of all soluble solutes completely dissolved in solution and is as follows:

H+ (aq) + Cl- (aq) + Na+ (aq) + OH- (aq) --> H2O (l) + Na+ (aq) + Cl- (aq)

3. The net ionic equation is the same as the total, or complete, ionic equation except that the spectator ions are left out. Spectator ions are those that appear on both the product side and the reactant side.

H+ (aq) + OH- (aq) --> H2O (l)

Also note that this is also an acid-base reaction.

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NaOH(aq)+HNO 3 (aq)?H 2 O(l)+NaNO 3 (aq)

NaOH -----------> Na+ +OH-
HNO3 --------- -> H+ + NO3-
--------------------------------------...
NaOH(aq)+HNO 3 (aq)----> (Na+ + NO3-) + H2O

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4. NaOH is a strong base and HCl a strong acid. Value of heat of neutralisation for a strong acid and a strong base is always 13.6 k.cal/mol or 57.3 kJ/mol.

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5.

Q = MCdT
Q = Heat in joules
M = Mass in grams
C = Specific Heat capacity
dT = Change in Temperature

From which M = Q /(CdT)

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