How Do You Calculate Energy Changes in Chemical Reactions?

[vg19]

Hi,

I having troubles with the following problems.

1) Calculate the energy change which will occur when 30g of nitrogen(II) oxide is converted to nitrogen(IV) oxide by reacting with oxygen in the air.

2)Copper(II) oxide can be reduced to copper metal and water by using hot hydrogen gas. Calculate the energy change when 54g of water is formed by this process.

I think my problem with these problems are I don't fully understand the concepts. If you could explain to me which formula I would use and why, I would really appreciate it.

Thanks so much

 

[mark-ashleigh]

Hello

Do you know how to do mass and energy balances? ...or even Hess's law?
Find a basic chemical engineering book and revise the chapter on energy balances...its easier then someone doing it for you ;)

If you can't find one or still have problems...do another post and i will help with the answers..

good luck

 

[wais]

Hi there,

I'm happy to help with your chemistry energy calculation problems. Let's take a look at each one and break down the steps to solve them.

1) To calculate the energy change for this reaction, we need to use the equation: ∆H = ∑∆H(products) - ∑∆H(reactants). This means that we need to calculate the sum of the enthalpies of the products and subtract the sum of the enthalpies of the reactants. First, we need to balance the chemical equation:

2NO + O2 → 2NO2

Next, we need to determine the enthalpy values for each substance. The enthalpy of formation for nitrogen(II) oxide is 90.25 kJ/mol and for nitrogen(IV) oxide is 33.2 kJ/mol. Since we have 30g of nitrogen(II) oxide, we need to convert it to moles by dividing by its molar mass (46g/mol). This gives us 0.65 moles of nitrogen(II) oxide. Similarly, we have 0.65 moles of nitrogen(IV) oxide. Therefore, the energy change for this reaction is:

∆H = (2 x 0.65 x 33.2) - (2 x 0.65 x 90.25) = -99.45 kJ

The negative sign indicates that this reaction is exothermic, meaning that energy is released.

2) For this problem, we need to use the same equation: ∆H = ∑∆H(products) - ∑∆H(reactants). The balanced chemical equation is:

CuO + H2 → Cu + H2O

The enthalpy of formation for copper(II) oxide is -156.6 kJ/mol and for water is -285.83 kJ/mol. We have 54g of water, which is equivalent to 3 moles (54/18 = 3). Therefore, the energy change for this reaction is:

∆H = (3 x -285.83) - (-156.6) = -572.49 kJ

Again, the negative sign indicates that this reaction is exothermic.

I hope this helps to clarify the concepts and steps involved in these types of calculations. Remember to always balance the chemical equation and use the correct enthalpy values for