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ALL CIE CHEMISTRY DOUBTS HERE !!

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$!$RatJumper$!$:
Thank you :)

moon:
plz can somebody help me in calculation  ppr2   Nov.07 Q1 (e)
thanx in advance.

$!$RatJumper$!$:

--- Quote from: moon on October 17, 2010, 02:37:21 pm ---plz can somebody help me in calculation  ppr2   Nov.07 Q1 (e)
thanx in advance.


--- End quote ---

The general equation for enthalpy change of formation is: (delta)H(formation) = Products - Reactants

Now, the information they have given us for the reaction, is all in (delta)H(combustion). So what you have to do, is simply reverse the signs on each of the values to change them to the formation values. In this case, since they are all negative, we make them positive.

Once done that, now identify your products and your reactants.
Products: C2H4
Reactants: 2C, 2H2

Then plug in their values for formation into the equation above like this:

(delta)H(formation) = (–1411.0) - [(2 * –285.9) + (2 * –393.7)]

Therefore, (delta)H(formation) = 51.8 kJ mol–1

P.S We multiplied them by 2 because the equation said 2C and 2H2. The 2 represents the moles so we take them into account.

Hope this helped

Deadly_king:

--- Quote from: $!$RatJumper$!$ on October 17, 2010, 03:08:28 pm ---The general equation for enthalpy change of formation is: (delta)H(formation) = Products - Reactants

Now, the information they have given us for the reaction, is all in (delta)H(combustion). So what you have to do, is simply reverse the signs on each of the values to change them to the formation values. In this case, since they are all negative, we make them positive.

Once done that, now identify your products and your reactants.
Products: C2H4
Reactants: 2C, 2H2

Then plug in their values for formation into the equation above like this:

(delta)H(formation) = (–1411.0) - [(2 * –285.9) + (2 * –393.7)]

Therefore, (delta)H(formation) = 51.8 kJ mol–1

P.S We multiplied them by 2 because the equation said 2C and 2H2. The 2 represents the moles so we take them into account.

Hope this helped

--- End quote ---

Good work buddy  ;)

+rep

moon:

--- Quote from: $!$RatJumper$!$ on October 17, 2010, 03:08:28 pm ---The general equation for enthalpy change of formation is: (delta)H(formation) = Products - Reactants

Now, the information they have given us for the reaction, is all in (delta)H(combustion). So what you have to do, is simply reverse the signs on each of the values to change them to the formation values. In this case, since they are all negative, we make them positive.

Once done that, now identify your products and your reactants.
Products: C2H4
Reactants: 2C, 2H2

Then plug in their values for formation into the equation above like this:

(delta)H(formation) = (–1411.0) - [(2 * –285.9) + (2 * –393.7)]

Therefore, (delta)H(formation) = 51.8 kJ mol–1

P.S We multiplied them by 2 because the equation said 2C and 2H2. The 2 represents the moles so we take them into account.

Hope this helped

--- End quote ---



I got the same answer -51.8 KJmol-1 but the markscheme's answer is +51.8KJmol-1 How come???

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