calculate h for the reaction 2c s h2 g a c2h2 g given c2 h2 g 52o2 g a 2co2 g h2o l h 12996 kjcs

Step 1: Write down the given reactions and their enthalpy changes ((Delta H)).

Calculate h for the reaction 2c s h2 g a c2h2 g given c2 h2...

Calculate ΔH for the reaction: 2C (s) + H2 (g) → C2H2 (g) Given: C2H2 (g) + 5/2O2 (g) → 2CO2 (g) + H2O (l) ΔH = -1299.6 KJ C (s) + O2 (g) → CO2 (g) ΔH = -393.5 KJ H2 (g) + 1/2 O2 (g) → H2O (l) ΔH = -285.5 KJ

Transcript

00:01 Hello everyone.

00:01 So for the given question, we need to calculate the enthalpy of formation for the given reaction.

00:06 So this is the reaction that we need to calculate the enthalpy formation.

00:09 And we have been given the enthalpy of formation for three reactions.

00:11 That is, first one is the ethythene molecule reacting with 5 by 2 moles of oxygen, giving us 2 moles of oxygen, giving us 2 more of water molecule.

00:20 The enthalpy of formation is given as minus 12, 99 .6 kilojoules per mole.

00:24 The second reaction is carbon reacting with oxygen, giving us carbon dioxide, enthalpya formation given as minus 393 .5.

00:30 The enthalpya formation is given as minus 393 .5.

00:30 Kilojoules per mole.

00:31 Third reaction is hydrogen reacting with half moles of oxygen giving us one mole of water.

00:36 The enthalpy of formation is given as minus 285 .5 kilojoules per mole.

00:41 So basically to find the enthalpy of formation for the above given reaction, we need to rearrange these three equations in such a way that two moles of carbon and one mole of hydrogen are on the reactant side and this ethyme, that is c2h2 gas, is on the product side.

00:54 So to do this, what we will do is first of all take the second equation, that is this one, and malicephemy, and malice to multiplied by 2.

01:02 So it will become 2 moles of carbon reacting with 2 moles of oxygen giving us 2 moles of carbon dioxide.

01:10 So since we have multiplied this equation by a factor of 2, so the enthalpy of formation will also be multiplied by a factor of 2.

01:17 So it will be 2 times minus 393 .5 kilojoules per mole.

01:25 So basically we have calculated the enthalpy of formation for reaction of 2 modes of carbon with 2 modes of oxygen giving us 2 moles of carbon dioxide right now in the given equation we have also the hydrogen on the reactant side so we'll take the third equation so we will add the third equation to this equation that is hydrogen reacting with half moles of oxygen giving us 1 mole of water molecule right so it doesn't help your formation for this reaction in the given question is minus 285 .5 kilojoules per mole now in the given reaction the methane gas is on the product side.

02:07 So basically we will take the first equation and inverse this equation.

02:11 So it will be two modes of carbon dioxide reacting with one mole of water, giving us one mole of this ethythine that is c2h2 and pi by two moles of this oxygen gas...

Question

Calculate ΔH for the reaction: 2C (s) + H2 (g) → C2H2 (g) Given: C2H2 (g) + 5/2O2 (g) → 2CO2 (g) + H2O (l) ΔH = -1299.6 KJ C (s) + O2 (g) → CO2 (g) ΔH = -393.5 KJ H2 (g) + 1/2 O2 (g) → H2O (l) ΔH = -285.5 KJ

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