I know that this is a repost, but I have a question:
HCl is a corrosive colourless gas that dissolves readily in water.
Aqueous HCL reacts with NaOH to form water and NaCl. In a simple
calorimeter, a 100.00 mL sample of 0.355 mol/L HCl(aq) is mixed
with 50.00 mL of excess NaOH(aq). During the reaction, there is a
rise in temperature by 4.200 °C. Calculate the molar enthalpy change
for the above reaction.
my question is about the volume multiplying the molarity to get moles, isn't it suppose to be multiplied by 0.150 L instead of the 0.100 L because " 100.00 mL sample of 0.355 mol/L HCl(aq) is mixed with 50.00 mL of excess NaOH(aq). " or is it just 0.100 L, and why is it 0.100L?
Also I have two other chemistry questions I need help on:
1.) Using the following reactions, calculate the heat of formation, ΔHf, of CS2 using Hess’ Law.
i. C(s) + O2(g) → CO2(g) ΔH = -393.3 kJ
ii. S(s) + O2(g) → SO2(g) ΔH = -293.72 kJ
iii. CS2(l) + 3 O2(g) → CO2(g) + 2 SO2(g) ΔH = -1108.76 kJ
My Answer:
ΔHsolution = ΔHproduct − ΔHreactant
ΔHreactant = ΔHproduct - ΔHsolution
ΔHCS2 = [ ΔHCO2 + 2ΔHSO2 ] - [ - ΔHsolution ]
ΔHCS2 = [ -393.3 + 2 ( -293.72 ) ] - [ - [ -1108.76]
ΔHCS2 = 128.02 KJ / ?
I put a question mark at the end because I don't know the last unit that is suppose to be there is moles or Kg? My friend got Kg, I don't how, but I think that it is suppose to be moles. Who is right and who is wrong?
2.) For the reaction below, initially 3.5 mol of NH3 are placed in a 4.0 L
reaction chamber. After 3.0 minutes only 1.6 moles of NH3 remain.
4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g)
a. Calculate the average rate of reaction with respect to NH3
b. Calculate the average rate at which H2O is being formed.
c. Calculate the average rate at which O2 is being consumed.
I don't know how to do b and c. I understand that once I find the answer to a that I can ratio the whole chemical equation to find the average rate for H2O and O2.
What I did to find a: (I don't know if it is right)
M1(NH3) = mol / L
= 3.5 mol / 4 L
= 0.875 mol/L
M2(NH3) = mol / L
= 1.6 mol / 4 L
= 0.4 mol/L
∆t = time
Rate of reaction = (M2 - M1) / (∆t2 - ∆t1)
= (0.4 mol/L - 0.875 mol/L) / ( 0 - 3 min)
= -0.475 mol/L / - 3 min
= 0.15833 mol⋅L^-1m^-1
I don't know if the answer is right, is it? Also once you get the final answer for a how would you ratio the chemical equation to answer b and c? can you please explain? Sorry for asking so many questions and thank you for help.
This is a little involved; therefore, let me answer these question one at a time in separate posts..
1. About the moles = M x L and the use of 0.1 or 0.150 for the volume. I wrote a lengthy explanation where you posed the same question when you first posted the question initially. Look there for that windy explanation, If you can't find it I can help and I'll send a link if you need it.
"1.) Using the following reactions, calculate the heat of formation, ΔHf, of CS2 using Hess’ Law.
i. C(s) + O2(g) → CO2(g) ΔH = -393.3 kJ
ii. S(s) + O2(g) → SO2(g) ΔH = -293.72 kJ
iii. CS2(l) + 3 O2(g) → CO2(g) + 2 SO2(g) ΔH = -1108.76 kJ
My Answer:
ΔHsolution = ΔHproduct − ΔHreactant
ΔHreactant = ΔHproduct - ΔHsolution
ΔHCS2 = [ ΔHCO2 + 2ΔHSO2 ] - [ - ΔHsolution ]
ΔHCS2 = [ -393.3 + 2 ( -293.72 ) ] - [ - [ -1108.76]
ΔHCS2 = 128.02 KJ / ?
I put a question mark at the end because I don't know the last unit that is suppose to be there is moles or Kg? My friend got Kg, I don't how, but I think that it is suppose to be moles. Who is right and who is wrong? "
I don't know where you picked up that delta H solution thing. It should be:
dHrxn = (n*dHf products) - (n*dHf reactants)
CS2(l) + 3 O2(g) → CO2(g) + 2 SO2(g) ΔH = -1108.76 kJ
dHrxn = (1*dHf CO2 + 2*dHf SO2) - (1*dHf CS2 + 3*dHf O2)
-1108.76 = (1*-393.3 + 2*-293.72) - (1*CS2 + 3*0)
Solve for CS2. All of the units are kJ (actually they are kJ/mol but the mol often is not written down) so the end unit is kJ/mol for dHf ^o for CS2.
Hope this helps.Follow up if needed.
"2.) For the reaction below, initially 3.5 mol of NH3 are placed in a 4.0 L
reaction chamber. After 3.0 minutes only 1.6 moles of NH3 remain.
4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g)
a. Calculate the average rate of reaction with respect to NH3
b. Calculate the average rate at which H2O is being formed.
c. Calculate the average rate at which O2 is being consumed.
I don't know how to do b and c. I understand that once I find the answer to a that I can ratio the whole chemical equation to find the average rate for H2O and O2.
What I did to find a: (I don't know if it is right)
M1(NH3) = mol / L
= 3.5 mol / 4 L
= 0.875 mol/L
M2(NH3) = mol / L
= 1.6 mol / 4 L
= 0.4 mol/L
∆t = time
Rate of reaction = (M2 - M1) / (∆t2 - ∆t1)
= (0.4 mol/L - 0.875 mol/L) / ( 0 - 3 min)
= -0.475 mol/L / - 3 min
= 0.15833 mol⋅L^-1m^-1
I don't know if the answer is right, is it? Also once you get the final answer for a how would you ratio the chemical equation to answer b and c? can you please explain? Sorry for asking so many questions and thank you for help".
I think your average calculation is correct although I think it is easier to do mols first but that just a matter of choice; i.e., I would have subtracted 3.5 mols - 1.6 mols = 1.9 and that divided by 4 L = 0.475 moles/L = M and 0.475/3 min = 0.158 M/L*s
There is some confusion, in my mind, about the use of reaction rate, reaction consumption, and rate of reaction, but I believe for the equation, the reaction rate for NH3 is 1/4*0.158 as the defined rate of reaction for NH3. Remember the negative sign for reactants and positive sign for formation of products.
Then the rate of disappearance of O2 is 5 times that; i.e., 0.158/4 x 5 = ?
etc. Hope this helps. Just let me know if I can help further. I don't mind the questions at all; in fact I prefer answering questions you have about the why and why not over just going through a calculation. Showing me what you've done and what you do/dont understand is very helpful in letting me know what you are having trouble with. Good luck in your chemistry studies. Stick with it.
oops. I see I made a typo on several units; e.g. I wrote 0.158 M/L*s when I should have typed in mole/L*s OR M/s. You'll see two or three like that.
What do you mean by "Remember the negative sign for reactants and positive sign for formation of products." how would this play out (would the outcome of the answer be negative, positive? I'm confused can you give an example and explain please)? Also, Why did you divide NH3 by 4? "NH3 is 1/4*0.158 as the defined rate of reaction for NH3." Can you elaborate?
To answer your first question:
In the given problem, you have a 100.00 mL sample of 0.355 mol/L HCl(aq) and you want to calculate the moles of HCl. To do this, you need to use the formula:
moles = volume (in liters) x molarity
Since the volume is given in milliliters, you need to convert it to liters by dividing by 1000:
volume = 100.00 mL = 100.00 mL / 1000 mL/L = 0.100 L
Now, you can calculate the moles of HCl:
moles = 0.100 L x 0.355 mol/L = 0.0355 mol
Therefore, the correct volume to use is 0.100 L.
Regarding your second question:
For the calculation of the heat of formation, ΔHf, of CS2 using Hess' Law, your approach is correct. The unit of ΔHf will depend on the units of the enthalpies of the reactions given. In this case, all the enthalpies are given in kJ/mol. Therefore, the unit of ΔHCS2 will also be kJ/mol. Your friend's answer of kg is incorrect.
For your third question:
To calculate the average rate of reaction with respect to NH3, you have correctly calculated the initial and final concentrations. The average rate of reaction is given by the change in concentration divided by the change in time:
Rate = (M2 - M1) / (∆t2 - ∆t1)
Using your values:
Rate = (0.4 mol/L - 0.875 mol/L) / (3 min - 0 min)
= -0.475 mol/L / 3 min
= -0.15833 mol⋅L^-1⋅min^-1
Therefore, the average rate of reaction with respect to NH3 is -0.15833 mol⋅L^-1⋅min^-1.
To calculate the average rate of formation of H2O, you would use the stoichiometric coefficients from the balanced equation. In this case, the ratio of NH3 to H2O is 4:6. So, the average rate of formation of H2O would be (4/6) times the average rate of reaction with respect to NH3.
Similarly, the average rate of consumption of O2 would be (5/4) times the average rate of reaction with respect to NH3.
I hope this helps! Let me know if you have any further questions.