Calculating Molar Heat/enthalpy questions (1 Viewer)

[clintmyster]

I got these questions and im totally stuck. Help would be much appreciated!


Calculate the amount of heat released or absorbed when:

a) 5g sulfur dixoide reacts with excess oxygen (delta H = -198 kJ/mol)
b) 25 mL 0.20 mol/L HCl is added to 50mL if a solution containing excess ammonia (delta H = -52 kJ/mol)

answers

a)7.7kJ released
b)0.26 kJ released

 

[midifile]

I got these questions and im totally stuck. Help would be much appreciated!


Calculate the amount of heat released or absorbed when:

a) 5g sulfur dixoide reacts with excess oxygen (delta H = -198 kJ/mol)
b) 25 mL 0.20 mol/L HCl is added to 50mL if a solution containing excess ammonia (delta H = -52 kJ/mol)

answers

a)7.7kJ released
b)0.26 kJ released

b. n(HCl) = cV(L)
= 0.2 x 0.025
= 0.05
delta H = 0.05 x -52
=-0.26 kJ

I think the answer for qu a is wrong and should be -15.45kJ

 

[clintmyster]

b. n(HCl) = cV(L)
= 0.2 x 0.025
= 0.05
delta H = 0.05 x -52
=-0.26 kJ

I think the answer for qu a is wrong and should be -15.45kJ

why do you multiply the delta H of .52 by 0.05 instead of dividing it? Im confused on that part.

 

[hamzie]

b. n(HCl) = cV(L)
= 0.2 x 0.025
= 0.05
delta H = 0.05 x -52
=-0.26 kJ

I think the answer for qu a is wrong and should be -15.45kJ

Yes i got that as well (15.45)
-------------------------------------

Delta H is in kj divide by mol

that is

kjmol-1

So delta H = H / Number of moles

MOLES OF SULFUr = 5g / 64g

H we have to find

DELTA H we are given

-198 = H / (5 divide by 64)

-198 x (5 divide by 64) = H

 

[midifile]

why do you multiply the delta H of .52 by 0.05 instead of dividing it? Im confused on that part.

do you mean -52?

If so, the molar enthalpy change is -52 kJ (1 mole). However you only had 0.05 moles so it is -52 x 0.05.

If the question said that 0.05 moles released 52kJ of energy then you would divide by 0.05 moles to find the molar enthalpy.

 

[hamzie]

Dont get confused by that chics working out

Moles of HCl = 0.2

delta H = -52

DELTA H = H / MOLES

-52 = H / (1 divide by 100)

-52 x (1 divide by 100) = H

 

[clintmyster]

yeah i understand now. Just dont know how to do one last question.

50mL 0.20mol/L lead Nitrate solution is added to excess potassium iodide contained in 30mL of solution. Both solutions are initiallly at 19.6 degrees. After mixing, the temperature rose to 22.2 degrees. Calculate the enthalpy change (per mole of Pb2+) for the reaction?

Answer is : -87kJ/mol

 

[tau281290]

Hey, I got 7.7 for Q a)

you just need to:

5/64/2 x 198

Remember there is excess oxygen and write the chemical equation out first.