AQA energetics help.

I'm so baffled on how to work out the multiple choice questions on this topic... Can someone please explain a way.

Many thanks.

Consider the reactions
C2H4(g) + 2O2(g) → 2CO(g) + 2H2O(g) ∆H = −758 kJ mol−1
2C(s) + 2H2(g) → C2H4(g) ∆H = +52 kJ mol−1
H2(g) + O2(g) → H2O(g) ∆H = −242 kJ mol−1

The enthalpy of formation of carbon monoxide is
A −111 kJ mol−1
B −163 kJ mol−1
C −222 kJ mol−1
D -464 kJ mol−1

(The answer is A apparently)

1. Write out the equation for the desired reaction
2. Using the four rules of number (+, -, x, divide) manipulate the three given equations to obtain the equation in 1.
3. Do the same to the energy value and you have the answer.

Ok so I did -242 x 2 to work out 2 moles of H2O then took that away from the ∆H of the first equation leaving me with -274.

The enthalpy of O2 and C is 0 so I'm left with the +52 to sub into the first equation...

Now I'm not sure on the final stage to get the answer.

you didn't follow the procedure that i gave you.

1. Where is the desired equation?

C + 1/2O2 --> CO

equation 1: C2H4(g) + 2O2(g) → 2CO(g) + 2H2O(g) ......... ∆H = −758 kJ mol−1

equation 2: 2C(s) + 2H2(g) → C2H4(g) ............... ∆H = +52 kJ mol−1

equation 3: H2(g) + 1/2O2(g) → H2O(g) ................ ∆H = −242 kJ mol−1

Right, so construct this equation by:

Dividing equation 2 by 2

equation 4: C(s) + H2(g) → 1/2C2H4(g) ........... ∆H = +26 kJ mol−1

Divide equation 1 by 2

equation 5: 1/2C2H4(g) + O2(g) → CO(g) + H2O(g) ......... ∆H = −379 kJ mol−1

add equation 4 and 5:

equation 6: C(s) + H2(g) + O2(g) → CO(g) + H2O(g)..... ∆H = −353 kJ mol−1
equation 3: H2(g) + 1/2 O2(g) → H2O(g) ................ ∆H = −242 kJ mol−1

Now subtract equation 3 from 6

C(s) + 1/2 O2(g) → CO(g)........ -353 --242 = -111 kJ mol^-1

Hi I've come across a similar question where I understand the process of working it out now however I'm a bit confused on the equation I'm supposed to form...

Using the data below, which is the correct value for the standard enthalpy of formation for TiCl4(l)?

equation 1: C(s) + TiO2(s) + 2Cl2(g) → TiCl4(l) + CO2(g) ∆H = −232 kJ mol−1
equation 2: Ti(s) + O2(g) → TiO2(s) = −912 kJ mol−1
equation 3: C(s) + O2(g) → CO2(g) = −394 kJ mol−1

A −1538 kJ mol−1
B −1094 kJ mol−1
C −750 kJ mol−1
D +286 kJ mol−1

I initially thought it was A as I thought the equation was Ti + 2Cl2 --> TiCl4 but the answer is C apparently. Then I realised isn't the equation just the first equation they have given?

Thanks

You have the equation correct.
You must construct it from the equations given:

I have removed the states for clarity.

start with equation 2:
add equation 1:


equation 1: C + TiO₂ + 2Cl₂ → TiCl4 + CO2 ........ ∆H = −232 kJ mol−1
equation 2: Ti + O₂ → TiO₂ ............ ∆H = −912 kJ mol−1
------------------------------------------------------------------add
equation 4: C + 2Cl₂ + Ti + O₂ → TiCl4 + CO2 ............ ∆H = −1144 kJ mol⁻¹

Now subtract equation 3: (or reverse equation 3 and add it)

equation 4: C + 2Cl₂ + Ti + O₂ → TiCl4 + CO2 ............ ∆H = −1144 kJ mol⁻¹
equation 3: C + O2 → CO2 = −394 kJ mol⁻¹
----------------------------------------------------------------- subtract

2Cl₂ + Ti → TiCl4 ........... ∆H = −1144 + 394kJ mol⁻¹


∆H = −750 kJ mol⁻¹