Can Anyone Answer These 10th Grade Questions That Deal With Calculating Enthalpy Changes

Can someone answer these 10TH GRADE ENTHALPY CHANGE questions?

1) Calculate ∆Hº for the reaction P4 + 10Cl2 --> 4PCl5 from the following reactions

P4 + 6Cl2 --> 4PCl3 ∆Hº= -1279 kJ
PCl3 + Cl2 --> PCl5 ∆Hº= -124 kJ

2) Calculate ∆H for the reaction NO + O -->NO2 from the following reactions.
NO + O3 -->NO2 + O2 ∆H= -198.9 kJ
2O3 --> 3O2 ∆H= -284.6
O2 --> 2O2 ∆H= 495 kJ

Thank you so so so much to those who answer...I will select a best answer.
Could you please explain how to do at least one of these problems? because I really don't understand where to even start with these types of problems...thank you so much

Can anyone answer these 10TH GRADE questions that deal with CALCULATING ENTHALPY CHANGES?

1. You want the addition of the two reactions to result in the given final reaction P4 + 10Cl2 --> 4PCl5
The step equations may have to be multiplied/divided and/or reversed to create the necessary canceling of substances to have the two step equations result in the final.

P4 + 6Cl2 --> 4PCl3 ∆Hº= -1279 kJ here we need all the substances except PCl3
PCl3 + Cl2 --> PCl5 ∆Hº= -124 kJ notice that for PCl3 to cancel out we need 4 of them in the second equation so it must be multiplied by 4.

P4 + 6Cl2 --> 4PCl3 ∆Hº= -1279 kJ
4PCl3 + 4Cl2 --> 4PCl5 ∆Hº= -496 kJ
Now we can cancel 4PCl3 because it appears on both sides.
Then we add the remaining substances and the ∆Hºs resulting in:
P4 + 10Cl2 --> 4PCl5 ∆Hº= -1775 kJ

2. Following the same procedure working with three step equations resulting in: NO + O -->NO2

NO + O3 -->NO2 + O2 ∆H= -198.9 kJ notice that NO2 is on the correct side so this equation does not need to be reversed. And notice that its one NO2 so no multiplication necessary.

2O3 --> 3O2 ∆H= -284.6 if reversed and divided by 2 will allow the cancellation of O3. This results in
1.5O2 --> O3 ∆H= + 284.6 the ∆H changes sign when the equation reverses.

O2 --> 2O ∆H= 495 kJ This equation needs to be reversed and divided by 2 resulting in
O --> 1/2 O2 ∆H= - 247.5 kJ

Now lets write the three equations one under the other for clarity:
1) NO + O3 -->NO2 + O2 ∆H= -198.9 kJ
2) 1.5O2 --> O3 ∆H= + 284.6
3) O --> 1/2 O2 ∆H= - 247.5 kJ

The O3s in equations 1) and 2) cancel. The O2s on the right in 1) and 3) cancel the O2 on the left in equation 2).

We are left with NO + O -->NO2 ∆H= (-198.9 + 284.6 - 247.5) = -161.8 kJ

ther you go!

Can someone answer this 10TH GRADE ENTHALPY CHANGE question?

If you payed any attention at all you'd know this is a Hess' law problem. You have to flip, multiply and add the equations to arrive at the end result. Look up Hess' Law on google and you'll find many sites that explain it in detail

H2 + 1/2O2 -----------------> H2O -286
2C + 2O2 ----------------------> 2CO2 -788
2CO2 + H2O ------------------> C2H2 + 5/2 O2 +1300
--------------------------------------...
2C + H2 ------------------------> C2H2 +226

Can anyone answer these 10TH GRADE questions that deal with CALCULATING HEATS OF REACTIONS?

I'll just help you with 1 and 3. You have to add the equations together. Switch the sign if you do the opposite (i.e. for the reaction of X+Y-->Z the opposite is Z-->X+Y). Multiply enthalpy by what you multiply the chemical equation by.
Pb(s) + 2Cl2(g) --> PbCl4(l) ∆H= -329.2 kJ
+
PbCl2(s) --> Pb(s) + Cl2(g) ∆H= 359.4 kJ (notice the sign change. in this case no multiplication is necessary.)

So
Pb(s) + 2Cl2(g) + PbCl2(s) --> Pb(s) + Cl2(g) + PbCl4(l)
Cancel out like compounds.
Cl2(g) + PbCl2(s) --> PbCl4(l)
So that proves it is the correct equation. Now add the enthalpies. (-329.2+359.4)= 30.2 kJ.
3. True cause the more negative the standard enthalpy of formation means its losing more energy to make the compound which also means that its gonna take more energy to break the compound and therefore, the compound is more stable.
Ok I'm too lazy to do the other one but good luck!

The enthalpy changes of the following reactions can be measured?

Your goal is to combine the given equations in some way so that when you add them together, you are left with the desired equation.

Start with the first equation:
C2H4(g) + 3 O2(g) --> 2 CO2(g) + 2 H2O (l) ΔH° = -1411.1 kJ

Flip the second equation over, changing the sign of its ΔH°:
2 CO2(g) + 3 H2O(l) --> C2H5OH(l) + 3 O2(g) ΔH° = +1367.5 kJ

WHen you add these together and simplify, you are left with the equation. Adding the ΔH° values gives:
C2H4(g) + H2O(l) --> C2H5OH(l) ΔH° = -1411.1 kJ + 1367.5 kJ = -43.6 kJ

Can someone answer these 10TH GRADE CHEM questions on HEAT OF COMBUSTION?

1. 4Al(s) + 3O2(g) --> 2Al2O3(s). 8.17 g Al / 27g/mol = 0.3036 mols

Ratio of Al : Al2O3 = 4 : 2 = 2 : 1 = 0.3036 : x ; x = 0.3036 / 2 = 0.1513 mols Al2O3
-1680 kJ / mol, x 0.1513 mols = -254.2 kJ heat released

2. Equation 1 H2(g) + Cl2(g) --> 2HCl(g) ∆H= -185 kJ, has 2HCl on the right but the final equation has 4HCl on the left so this equation has to be reversed and multiplied by 2:
Eq. 1 4HCl ----> 2H2 + 2 Cl2 ∆H= +370 kJ Note that ∆H is no positive because of the reversal.
Equation 2 2H2(g) + O2(g) --> 2H2O(g) ∆H= -483.7 kJ. Has all components on the correct sides and the proper amounts so that we can now add the two equations together to get the final equation.

Starting with revised Eq. 1, we bring down 4HCl on the left and from Eq. 2 + O2. The 2H2 on the right in Eq 1 and on the left in Eq. 2 cancel each other out. That leaves 2Cl2 on the right in Eq. 1 and 2H2O on the right in Eq. 2 to bring down to the final equation resulting in the desired result:
4HCl(g) + O2(g) --> 2Cl2(g) + 2H2O(g)

Ta da!

Why do you use the limiting reactant to determine enthalpy change?

If there is no reaction, delta H would be zero, right? The limiting reactant determines how much chemicals can be reacted, no matter how much the abundant reactants there are---the excess part are not reacted, thus does not have any contribution to delta H.

The heat of combustion of propane and calculating the heat of formation...?

First write the equation of the combustion reaction:

C3H8(g) + 5O2(g) ---> 3CO2(g) + 4H2O(l) ...dH = -2220.1 kJ/mol

Delta H subscript (f) means "the enthalpy of formation" of the substance under the standard state conditions (25 C and 1 atm).

Enthalpies of formations of gaseous CO2 and liquid water are given. Heat of combustion (nethalpy of combustion) of propane is also known, then;
delta Hf for C3H8 can be calculated from the enthalpy of the reaction. By definition; the enthalpy of a reaction (delta Hr);
delta Hr = Total delta Hf (products) - Total delta Hf (reactants)

Note: The enthalpy of formation of the most stable form of the elements is zero.

dH(combustion) = [3dHf (CO2) + 4dHf (H2O)] - [5dHf (O2) + dHf (C3H8)]

-2220.1 = [3(-393.5) + 4(-285.3)] - [ 0 + dHf (C3H8)]

-2220.1 = [-1180.5) - 1141.2] - dHf (C3H8)
-2220.1 = - 2321.7 - dHf (C3H8)
dHf (C3H8) = 2220.1 - 2321.7 = - 101.6 kJ/mol