Unit 4 questions

[bec]

this is a pretty basic question...but:

Many major car makers have unveiled hydrogen-powered cars. In the engines of these cars, hydrogen reacts with oxygen from the air to produce water.
2H2(g) + O2(g) -->  2H2O(g)

The magnitude of the activation energy of this reaction is 1370
H for this reaction is -572

Several groups of scientists have claimed to have split water into hydrogen and oxygen using a molybdenum catalyst:
2H2O(g)   2H2(g) + O2(g)

What is the value of H for this water-splitting equation?

The answer is 572 - why is that? Is it because 572kJ were released into the atmosphere when H20 was formed in the first eqn, so we need to re-add that 572 in order for the reverse reaction to work?

thanks!

[Mao]

basically right

its the energy stored in the bonds. basically the potential within the O=O bond is a lost stronger than the H-O-H, so when the double bond is broken, energy is released [because its lower energy level]

if you want to restore the double bond, you have to input some energy so it can regain its original strength

[coblin... where art thou and thy pwnful explanation?]

[bec]

thanks mao... that was what, 2 minutes?

[Mao]

00:04:02

[bec]

Q:
Hydrogen reacts explosively to form water.
Explain how the energy changes during bond-breaking and bond-forming affect the energy change for the reaction.

A: The energy change for the reaction is the difference between the energy absorbed to break the bonds in the H2 and O2 reactants, and the energy released when the bonds in the H2O product are made.

Isn't the energy change (which i'm assuming is the same as enthalpy change, H) the difference between the energy of the reactants before the bonds are broken, and the energy of the products once the new bonds have been formed? eg, if you look at a graph here, the H is he dif in energy between reactants and products, not from the peak of the graph to the lowest point:
http://images.google.com.au/imgres?imgurl=http://www.chem.tamu.edu/class/majors/tutorialnotefiles/endoea.gif&imgrefurl=http://www.chem.tamu.edu/class/majors/tutorialnotefiles/activation.htm&h=234&w=303&sz=3&hl=en&start=13&um=1&tbnid=WkefATvqD_nwVM:&tbnh=90&tbnw=116&prev=/images%3Fq%3Denthalpy%2Bchange%2Bexothermic%26um%3D1%26hl%3Den%26rlz%3D1T4GGIH_en-GBAU259AU259%26sa%3DN

thanks!

also, can someone explain how to do links so you don't need to display the whole entire website address?

[Mao]

yeah

you are saying the same thing

, which is the difference in energy levels

basically, it is how far up you have travelled up the peak minus how much drop, so they are really the same thing.

[btw someone put some light as to what supposed to be? i thought it is how much energy that need to be absorbed i.e. from reactant to peak, rather than the higher plateau of the two to the peak...]

\Edit: just had confirmation:
the graph for the endothermic reaction [first graph] is wrong. should extend from the base all the way to the peak.

[lanvins]

Can you just clear some things up for me;

Natural gas begins to burn when lit with a match. Why does it continue to burn when the match is taken away? (i think it has something to do with activation energy?)

Why must the energy absorbed, when the bonds break, be less than the energy released when new bonds form in the combustion of methane; which results in an overall release of thermal energy?

Why does a small increase in temperature have a greater impact on the rate than an increase at higher temperatures?

[Mao]

Natural gas begins to burn when lit with a match. Why does it continue to burn when the match is taken away? (i think it has something to do with activation energy?)

the reaction is exothermic, and the energy released is higher than the activation energy, so subsequent reaction is sustained.

Why must the energy absorbed, when the bonds break, be less than the energy released when new bonds form in the combustion of methane; which results in an overall release of thermal energy?

from previous, the reaction is sustained, hence is greater than , i.e. the energy absorbed [activation energy] is less than the energy released []

[lanvins]

could question two also have something to do with the the reactants having energy already and then absorbing extra energy, and thats why when the bonds form the the energy release is greater then the energy absorbed? ( i hope that makes sense)

[bec]

I just want to clear up something with equilibriums, K values and reaction quotients...is this stuff right?

- when a reaction is in equilibrium, that just means that the concentration of reactants and products has levelled out and doesn't change - even though the reaction is dynamic, the rate of the fwd and reverse reactions are the same
- if a reaction is in equilibrium, that doesn't meant that the yield is 100%, it just means that the yield is as high as it can possibly be
- reaction quotient is another term meaning concentration fraction
- equilibrium constant = K = the concentration fraction at the equilibrium
- if , there is roughly the same amount of products and reactants at the equilibrium

Also, is there a range of reaction quotients that shows a reaction is in equilibrium?
eg. if a reaction with reactants in particular concentrations gives the reaction quotient of 0.077, and you know that the K-value at the same temperature is 0.020, that shows that the concentration of products needs to increase for it to reach equilibrium...but what if the reaction quotient was 0.018 and K=0.02 - is there ever a case of "close enough is good enough" since it's all calculated experimentally?

thank you!

[Collin Li]

- if a reaction is in equilibrium, that doesn't meant that the yield is 100%, it just means that the yield is as high as it can possibly be

If a reaction is in equilibrium, it doesn't mean the yield is 100%. True.

It just means the yield is as high as it can be? Wrong. Yields can be increased with tactics such as creating an open system, which means the equilibrium may never be reached, but due to the driving force of the system trying to reach equilibrium, reactions can go to completion.

Also, is there a range of reaction quotients that shows a reaction is in equilibrium?
eg. if a reaction with reactants in particular concentrations gives the reaction quotient of 0.077, and you know that the K-value at the same temperature is 0.020, that shows that the concentration of products needs to increase for it to reach equilibrium...but what if the reaction quotient was 0.018 and K=0.02 - is there ever a case of "close enough is good enough" since it's all calculated experimentally?

No -- maybe in the labs some practice like that goes on, but you're just supposed to use reaction quotients in theory: Q < K means forward reaction, Q > K means backwards reaction.

[bec]

thanks for that coblin

i don't really understand this:

It just means the yield is as high as it can be? Wrong. Yields can be increased with tactics such as creating an open system, which means the equilibrium may never be reached, but due to the driving force of the system trying to reach equilibrium, reactions can go to completion.

what's an open system?

[Collin Li]

Open system is when you don't bottle in all of the reactants and products.

I can let my CO2 gas fly off into the atmosphere for example, and that is basically a continuous removal of a product, which would drive the reaction to the right until there's no more reactant.

[bec]

got it, thank you!

[bec]

Why does the equilibrium constant of an endothermic reaction decrease towards a vertical "asymptote" (inverted commas beause there isn't really an asymp, it just sort of looks like one)? I'm looking at a graph that my teacher gave us, and the chart for K vs temp in exothermic reactions decreases towards a horizontal "asymptote", which makes sense, and I don't understand why the exothermic would be curved in the other direction.

Sorry that question barely even makes sense to me, but if anyone understands what I mean that would be greatly appreciated. Otherwise I'll take a photo of it...

Thanks!