Group II metal carbonates MCO3, can undergo many types of reaction. One reaction is an acid-base reaction with HCl acid according to the equation
MCO3 (aq) + 2HCl(aq) -> MCl2 (aq) + CO2 (g) + H2O (l) enthalpy change (standard) = -61.5 kJ/mol
An experiment was set up to determine the Mr of MCO3. (a diagram showing a styrofoam cup with a thermometer was drawn)
50cm3 of x g dm-3 MCO3 was added to 60cm3 of 2 mol dm-3 HCl in a polystyrene cup of negligible heat capacity. MCO3 was known to be the limiting reagent.
(a) State all the data u would collect in the expt
I wrote the average initial temperature (dont think theres anything else, pls advise if there is a need to improve this ans)
(b) Use algebraic terms, where appropriate, to represent the data obtained from your experiment. (eg T for temperature reading and m for mass reading) Outline how these data can be used to determine the Mr of MCO3. Express your answer in terms of x. You many assume the specific heat capacity of water to be 4.2J g-1 K-1.
I believe this is more of an energetics question. But no matter how I try to solve it, I simply cant express this in term of x. Please explain to me.
Thanks.
Another question.
Solid SrCO3 undergoes thermal decomposition accodring to the eqn shown
SrCO3 (s)-> SrO (s) + CO2 (g)
given that enthalpy, delta H = +234.6kJ/mol and delta S = +171 J/mol/K... was asked to calculate min. temperature. The part i am unsure of comes in here: The actual temperature at which the solid decomposes is lower than that calculated. Explain why this is so.
Thanks again
Ur 1st qn :
Using Q = MCdeltaT and using Enthalpy change = Q divided by no. of moles of limiting reactant, u can determine the no. of moles of MCO3 (limiting reactant) present (in terms of algebraic variables T and m).
U can also find the sample mass of MCO3 used (in terms of x), since u're given the mass conc.
Mr or molar mass = sample mass / no. of moles.
Ur 2nd qn :
The discrepancy between calculated value and actual value, is due to the false assumption used in ur calculations that enthalpy and entropy change values do not change significantly with temperature.
Thanks so much UltimaOnline! Appreciate the help.
Hi all, I have 2 more questions to seek help for.
1. Hydrazine, with a density of 1.0g ccm-3 has a lower density than H2O2 whose density is 1.4g cm-3. Account for this difference.
2. 0.0714mol of gaseous C3H6 was placed in a closed vessel of volume 10.0 dm3 at 450K. The sytem was then allowed to come to a state of equilibrium at a constant temperature of 450K.
Calculate the initial pressure of C3H6, assuming ideal gas behaviour. I got 26700Pa.
Then I was asked to estimate the value of Kp of this reaction 2 C3H6 -><- C2H4 (g) + C4H8(g) which I got 54.6. (from a log graph of ln Kp against temperature in Kelvin)
The main question comes here: Calculate the %tage yield of C2H4 under these conditions.
Thanks once again
Originally posted by Audi:Thanks so much UltimaOnline! Appreciate the help.
Hi all, I have 2 more questions to seek help for.
1. Hydrazine, with a density of 1.0g ccm-3 has a lower density than H2O2 whose density is 1.4g cm-3. Account for this difference.
2. 0.0714mol of gaseous C3H6 was placed in a closed vessel of volume 10.0 dm3 at 450K. The sytem was then allowed to come to a state of equilibrium at a constant temperature of 450K.
Calculate the initial pressure of C3H6, assuming ideal gas behaviour. I got 26700Pa.
Then I was asked to estimate the value of Kp of this reaction 2 C3H6 -><- C2H4 (g) + C4H8(g) which I got 54.6. (from a log graph of ln Kp against temperature in Kelvin)
The main question comes here: Calculate the %tage yield of C2H4 under these conditions.
Thanks once again
Q1. Molar mass of H2O2 > Molar mass of N2H4; N2H4 is larger than H2O2; hydrogen bonding in H2O2 slightly stronger than in N2H4.
Q2. Do an ICE table, and use Kp value to solve for the equilibrium molarity or partial pressure of C2H4. Apply stoichiometry to calculate the theoretical molarity or partial pressure of C2H4 if the reaction went to completion (ie. non-equilibrium). Hence calculate % yield (ie. actual / theoretical).
Ure welcome, Audi.