## Title

4 posts / 0 new
chemistrysubjec...
Offline
Joined: 10/06/2007 - 22:31
Posts: 347
chemistrysubjec...
Offline
Joined: 10/06/2007 - 22:31
Posts: 347

How do you calculate reating masses?

I would use moles. for instance if you have 24 g of carbon and burn it, the equation is
C + O2 > CO2 (sorry can't do superscript and subscripts in this package).

RAM of carbon is 12 so 24 g is two moles.

Equation tells you one mole of carbon goes with one mole of oxygen to make one mole of carbon dioxide.

SO, two moles of carbon will make two moles of carbon dioxide. One mole of carbon dioxide is 44 g (Mr is 12 + 16 + 16).

SO, two moles of carbon dioxide has a mass of 88 g.

Energy cycles

Try to tell the story of the reaction.

methane + oxygen > carbon dioxide +water
CH4 + 2 O2 > CO2 + 2 H2O

If you start from the elements in their standard states and use them to make the substances in the equation (these are called the enthalpy change of formation), the difference between the energy given out by making the (methane and oxygen) and making the (carbon dioxide and water) is the energy change during the reaction.

Elements to CH4 and 2 O2 = X

Elements to CO2 and 2 H2O = Y

Energy change = Y - X

Watch out for the multipliers (since there are two mkoles of oxygen and two moles of carbon dioxide) and watchgnout for the negative numbers.

Alternatively you can do it by bond energy calculation.

Find the total energy required to break the bonds on the LHS. Call this A.

Find the total energy released when the bonds are formed on the RHS. Call this B.

Energy change in the reaction is A - B. If the energy released by the bonds being formed on RHS is larger than the energy required to break the bonds on the LHS, the reaction is exothermic and the sign for the energy change is negative.

Hess's Law says that the energy change during a reaction is the same no matter which route the reaction takes.

In other words, then enrgy change for a two step reaction is the same as the energy change for each of the steps added together. Like all the best laws, it is common sense.

Link between equilibrium data and kinetic data.

A bit complicated but....

At equilibrium the rate of the forward reaction = the rate of the backward reaction.

As an example,for the equilibrium between
A + B and C + D
Rate forward = kf x [A] x [B]
Rate backward = kb x [C] x [D]

If both rates are the same,
kf x [A] x [B]=kb x [C] x [D]

SO if we get both the rate constants together,

kf / kb = {[C] x [D]} / {[A] x [B]}

If we call the ratio of the rate constants "Kc", we have the well known expression for equilibrium constant.

Hope this helps. It is difficult to do this very quickly but I know you are in a rush for your exam.

If it helps, great. If not, come back on the site and explain step by step what you find difficult (and which exam you are heading for) and I will try to help you more slowly and carefully.

Emily222115
Offline
Joined: 08/01/2009 - 17:39
Posts: 2

Thank - you very much!

It was kind of you to spend time helping me.

I understood everything and it helped me very much in my exam.

:D

chemistrysubjec...
Offline
Joined: 10/06/2007 - 22:31
Posts: 347

Glad it helped. If there is anything else that you need help with, please get back online and I will try to help you.