Balanced Reaction Equations (3 of these): Data: 1. Concentration of HCI K M 2. C

Question

Balanced Reaction Equations (3 of these): Data: 1. Concentration of HCI K M 2. Concentration of NaOH d 3. Concentration of NH, 2 OOM 4. Concentration of NH ci O M Reaction 1 Reaction2 and HCNao1 ya. Reaction 3 HCI and NH NaOH and NH CI Volume of acid (mL) 0.0 40.0 Volume of base (mL) 50.O 40.0 Total volume (mL) Mass of total solution. (g) Final temp. (C) Initial temp. (C) AT (C) heat, q (J) AH (kj/mol) 2 163 5103935.0 initial temp. ()24.34. 5.0 40.505 4L004.3 Observations: record any and all relevant observations: Reagent Table

Explanation / Answer

For trial 1:

calculate the moles that are reacting

moles = Molarity * volume

moles NaOH = 2.05 * 0.0491 = 0.1

moles of HCl = 2 * 0.05 = 0.1

so you had 0.1 moles of acid and base, you have already calculated the heat generated, to calculate the enthalpy of neutralization you are supposed to divide by the moles of water generated, since the neutralization of NaOH and HCl is 1 to 1

NaOH + HCl = == H2O + NaCl, you will only get 0.1 moles of water so

molar enthalpy is = 6401.5 / 0.1 = 64015 J / mol, divide this to get KJ

molar enthalpy is 64.015 KJ / mol

Trial 2

The reaction is

NaOH + NH4Cl ===== NaCl + NH3 + H2O, as you can see you have a 1:1 reaction lets get the moles that are reacting

the moles of NaOH are

0.05 * 2.05 = 0.1025 moles

moles of NH4Cl

0.0499 * 2 = 0.0998 moles

lets make the calculations with 0.1 moles , so according to the reaction you could only be able to form just 0.1 moles of water, again divide the heat you got by this number of moles

Molar enthalpy is 474.05 / 0.1 = 4740.5 Joules / moles, divide this by 1000 to get

4.740 KJ / mol

For trial 3

HCl + NH3 ==== NH4Cl

if you repeat the calculation you will find that the moles are 0.0998 and 0.0997 moles, so we can make ccalculations with 0.1 moles, so we can say that 0.1 moles of NH4Cl are being created

We can calculate the molar enthalpy, for this case lets divide by the moles of NH4Cl  

Molar enthalpy = 4654.3 J / 0.1 moles = 46543 J / moles, divide by 1000 to get KJ

Molar enthalpy = 46.543 KJ / moles

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