Question

I have a couple questions the first one is:

1. Combine equations 1,2,3 to obtain equation 4. Then provide an equation for calculating the heat of reaction in kJ/mol.

Equation 1: MgO(s)+2HCl(aq)--->MgCl2(aq)+H2O(l)

Equation 2: Mg(s)+2HCl(aq)---->MgCl2(aq)+H2(g)

Equation 3: H2(g)+1/2O2(g)---->H2O(l)

2. Use Hess's Law to calculate the heat of reaction for the decompostion of methanol:

2CH3OH(l)--->2CH4(g)+O2(g)      Change Heat of reaction =?

Using this information:

H2O(g)--->H2(g)+1/2O2(g)            Change Heat= +241.82 kJ

CH4(g)+H2O(g)---->CO(g)+3H2(g) Change Heat = +206.10 kJ

2H2(g)+CO(g)---->CH3OH(l) Change Heat = -128.33 kJ

Maybe my assignment is poorly worded but I am confused as what to do, thank you for any help.

Answer 1

While solving problems related to thermochemical equation, the following facts are helpful.

a.When a thermochemical equation is multiplied by a factor 'n', the heat change also needs to be multiplied by 'n'

b.When a thermochemical equation is reversed, the sign of heat change for the reaction is also reversed.

c. When we add or subtract two thermochemical reaction, their respeective heat of reaction is also added or subtracted.

Let's first solve the second question

2. The given thermochemical reactions are

H2O(g)--->H2(g)+1/2O2(g) Heat Change = +241.82 kJ -------- (1)

CH4(g)+H2O(g)---->CO(g)+3H2(g) Heat Change= +206.10 kJ --------- (2)

2H2(g)+CO(g)---->CH3OH(l) Heat Change = -128.33 kJ - ------- (3)

Now we need  the heat of reaction for the decompostion of methanol as

2CH3OH(l)--->2CH4(g)+O2(g)

Since we have 2CH3OH(l) on reactant side, we need to reverse equation (3) and multiply by 2 and the new equation obtained is

2CH3OH(l) -----> 4H2(g)+2CO(g) , Heat Change = -2x(-128.33 kJ) = 256.66 KJ ------(4)

We need 2CH4(g) on right, hence  we need to reverse equation (3) and multiply by 2 and the new equation obtained is

2CO(g)+6H2(g) ------> CH4(g)+H2O(g), Heat Change = -2x(+206.10 kJ) = - 412.20 KJ -------(5)

Also we need one O2(g), hence ned to multiply equation (1) by 2 and new equation obtained is

2H2O(g) ------>2H2(g) + O2(g)            Heat Change = 2x(+241.82 kJ) = 483.64 KJ -------- (6)

Now on adding the new equations (4), (5) and (6), CO(g), H2(g) and H2O(g) will cancel out and we get the required equation which is

2CH3OH(l)--->2CH4(g)+O2(g) Heat Change = 256.66 KJ + (- 412.20 KJ) + 483.64 KJ = 328.10 KJ

Hence  heat of reaction for the decompostion of methanol = 328.10 KJ (answer)

1. In the same manner we can calculate heat change 4 th equation

Mg(s) + 1/2O2(g)--->MgO(s) --------(4)

The given equations are

Equation 1: MgO(s)+2HCl(aq)--->MgCl2(aq)+H2O(l) , DelH1

Equation 2: Mg(s)+2HCl(aq)---->MgCl2(aq)+H2(g) , DelH2

Equation 3: H2(g)+1/2O2(g)---->H2O(l) , DelH3

Since in equation (4),  Mg(s) and 1/2O2(g) are on left, we need to add equation (2) and (3). Also we need MgO(s) on right, we need to subtract equation (1)

Hence equation (2) + (3) - (1) =>

Mg(s) + 1/2O2(g)--->MgO(s) , DelH = (DelH2) + (DelH3) - (DelH1) -------- (4) (answer)