Chemistry_Grade_10-12 (1).pdf

The equilibrium for the reaction lies far to the left

Info icon This preview shows pages 320–323. Sign up to view the full content.

View Full Document Right Arrow Icon
are more reactants than products and therefore the yield is low. The equilibrium for the reaction lies far to the left. Important: Calculations made easy When you are busy with calculations that involve the equilibrium constant, the following tips may help: 1. Make sure that you always read the question carefully to be sure of what you are being asked to calculate. If the equilibrium constant is involved, make sure that the concentrations you use are the concentrations at equilibrium , and not the concentrations or quantities that are present at some other time in the reaction. 2. When you are doing more complicated calculations, it sometimes helps to draw up a table like the one below and fill in the mole values that you know or those you can calculate. This will give you a clear picture of what is happening in the reaction and will make sure that you use the right values in your calculations. Reactant 1 Reactant 2 Product 1 Start of reaction Used up Produced Equilibrium Worked Example 77: Calculating k c Question: For the reaction: SO 2 ( g ) + NO 2 ( g ) NO ( g ) + SO 3 ( g ) 306
Image of page 320

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
CHAPTER 16. REACTION RATES - GRADE 12 16.7 the concentration of the reagents is as follows: [SO 3 ] = 0.2 mol.dm 3 [NO 2 ] = 0.1 mol.dm 3 [NO] = 0.4 mol.dm 3 [SO 2 ] = 0.2 mol.dm 3 Calculate the value of k c . Answer Step 1 : Write the equation for k c k c = [ NO ][ SO 3 ] [ SO 2 ][ NO 2 ] Step 2 : Fill in the values you know for this equation and calculate k c k c = (0 . 4 × 0 . 2) (0 . 2 × 0 . 1) = 4 Worked Example 78: Calculating reagent concentration Question: For the reaction: S ( s ) + O 2 ( g ) SO 2 ( g ) 1. Write an equation for the equilibrium constant. 2. Calculate the equilibrium concentration of O 2 if Kc=6 and [ SO 2 ] = 3 mol.dm 3 at equilibrium. Answer Step 1 : Write the equation for k c k c = [ SO 2 ] [ O 2 ] (Sulfur is left out of the equation because it is a solid and its concentration stays constant during the reaction) Step 2 : Re-arrange the equation so that oxygen is on its own on one side of the equation [ O 2 ] = [ SO 2 ] k c Step 3 : Fill in the values you know and calculate [O 2 ] [ O 2 ] = 3 mol.dm 3 6 = 0 . 5 mol.dm 3 Worked Example 79: Equilibrium calculations Question: Initially 1.4 moles of NH 3 (g) is introduced into a sealed 2.0 dm 3 reaction vessel. The ammonia decomposes when the temperature is increased to 600K and reaches equilibrium as follows: 307
Image of page 321
16.7 CHAPTER 16. REACTION RATES - GRADE 12 2 NH 3 ( g ) N 2 ( g ) + 3 H 2 ( g ) When the equilibrium mixture is analysed, the concentration of NH 3 (g) is 0.3 mol.dm 3 1. Calculate the concentration of N 2 (g) and H 2 (g) in the equilibrium mixture. 2. Calculate the equilibrium constant for the reaction at 900 K. Answer Step 1 : Calculate the number of moles of NH 3 at equilibrium. c = n V Therefore, n = c × V = 0 . 3 × 2 = 0 . 6 mol Step 2 : Calculate the number of moles of ammonia that react (are ’used up’) in the reaction. Moles used up = 1.4 - 0.6 = 0.8 moles Step 3 : Calculate the number of moles of product that are formed.
Image of page 322

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 323
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern