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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)

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)

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#2

(Original post by

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)

**sushma_roberts**)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)

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.

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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.

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.

(Original post by

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.

**charco**)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.

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#4

(Original post by

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.

**sushma_roberts**)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.

1. Where is the desired equation?

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C + 1/2 o2 --> co

(Original post by

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

1. Where is the desired equation?

**charco**)you didn't follow the procedure that i gave you.

1. Where is the desired equation?

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#6

(Original post by

C + 1/2O2 --> CO

**sushma_roberts**)C + 1/2O2 --> CO

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}

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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)?

C(s) + TiO2(s) + 2Cl2(g) → TiCl4(l) + CO2(g) ∆H = −232 kJ mol−1

Ti(s) + O2(g) → TiO2(s) = −912 kJ mol−1

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

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

C(s) + TiO2(s) + 2Cl2(g) → TiCl4(l) + CO2(g) ∆H = −232 kJ mol−1

Ti(s) + O2(g) → TiO2(s) = −912 kJ mol−1

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

(Original post by

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

**charco**)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}
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#8

(Original post by

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

**sushma_roberts**)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 must construct it from the equations given:

I have removed the states for clarity.

start with equation 2:

add equation 1:

equation 1: C + TiO

_{2}+ 2Cl

_{2}→ TiCl4 + CO2 ........ ∆H = −232 kJ mol−1

equation 2: Ti + O

_{2}→ TiO

_{2}............ ∆H = −912 kJ mol−1

------------------------------------------------------------------add

equation 4: C + 2Cl

_{2}+ Ti + O

_{2}→ TiCl4 + CO2 ............ ∆H = −1144 kJ mol

^{−1}

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

equation 4: C + 2Cl

_{2}+ Ti + O

_{2}→ TiCl4 + CO2 ............ ∆H = −1144 kJ mol

^{−1}

equation 3: C + O2 → CO2 = −394 kJ mol

^{−1}

----------------------------------------------------------------- subtract

2Cl

_{2}+ Ti → TiCl4 ........... ∆H = −1144 + 394kJ mol

^{−1}

∆H = −750 kJ mol

^{−1}

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Ah thanks so much!

(Original post by

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 + TiO

equation 2: Ti + O

------------------------------------------------------------------add

equation 4: C + 2Cl

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

equation 4: C + 2Cl

equation 3: C + O2 → CO2 = −394 kJ mol

----------------------------------------------------------------- subtract

2Cl

∆H = −750 kJ mol

**charco**)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 + TiO

_{2}+ 2Cl_{2}→ TiCl4 + CO2 ........ ∆H = −232 kJ mol−1equation 2: Ti + O

_{2}→ TiO_{2}............ ∆H = −912 kJ mol−1------------------------------------------------------------------add

equation 4: C + 2Cl

_{2}+ Ti + O_{2}→ TiCl4 + CO2 ............ ∆H = −1144 kJ mol^{−1}Now subtract equation 3: (or reverse equation 3 and add it)

equation 4: C + 2Cl

_{2}+ Ti + O_{2}→ TiCl4 + CO2 ............ ∆H = −1144 kJ mol^{−1}equation 3: C + O2 → CO2 = −394 kJ mol

^{−1}----------------------------------------------------------------- subtract

2Cl

_{2}+ Ti → TiCl4 ........... ∆H = −1144 + 394kJ mol^{−1}∆H = −750 kJ mol

^{−1}
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