100 mo lofh2swasputina100lvesselandheatedto1132

Info iconThis preview shows page 1. Sign up to view the full content.

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

Unformatted text preview: L = 0.0613 M 1.84 mo l/10.00 L = 0.184 M 0.387 mo l/10.00 L = 0.0387 M 0.387 mo l/10.00 L = 0.0387 M [H 2 O] [CH 4 ] (0.387 M)(0.387 M) = = 3.93 3 [CO] [H 2 ]3 (0.0613 M(0.184 M) Example 2: An 8.00 L reaction vessel at 491 °C contained 0.650 mo l H2, 0.275 mol I2 and 3.00 mo l HI. Assuming that the substances are at equilibrium, find the value of Kc at 491 °C. The equilibrium is: ¾¾ ® H2(g) + I2(g) ¬¾ 2HI(g) ¾ Moles at Equilibrium: 0.650 0.275 Molarit ies at Equilibrium: (moles/L) 0.650 mo les ¾¾¾¾¾ 8.00 L 0.275 mo les ¾¾¾¾¾ 8.00 L Molarities at Equilibrium: 0.08125 M 0.03438 M 3.00 3.00 mo les ¾¾¾¾¾ 8.00 L 0.375 M 2 [HI]2 (0.375 M) K c = = = 50.3 [H 2 ] [I2 ] (0.08125 M)(0.03438 M) 7 Chemistry 102 Chapter 14 Example 3: ¾¾ ® H2S, a colorless gas dissociates on heating: 2 H2S ¬¾ 2 H2 (g) + S2 (g) ¾ When 0.100 mo l of H2S was put in a 10.0 L vessel and heated to 1132 °C, it gave an equilibrium mixture containing 0.0285 mo l H2. What is the value o f Kc at this temperature? Note: It is conveni...
View Full Document

This document was uploaded on 03/18/2014 for the course CHEM 102 at Los Angeles Mission College.

Ask a homework question - tutors are online