CHEM 1100 table Secure | https://www.awh.aleks. si.exe/io.u-IgNslkr7 8P3jH-UIT L

Question

CHEM 1100 table Secure | https://www.awh.aleks. si.exe/io.u-IgNslkr7 8P3jH-UIT LCmFKBGScXhEWOiLTrFEOE O ADVANCED MA -Calculating equilibrium composition from an equilibrium constant Suppose a 500. mL flask is filled with 0.20 mol of H2 and 1.3 mol of HI. The following reaction becomes possible: H2(g) +12g 2HI(g The equilibrium constant K for this reaction is 0.387 at the temperature of the flask. Calculate the equilibrium molarity of 12. Round your answer to two decimal places. Explanation Check MacBook Air

Explanation / Answer

Volume of the flask is 500 mL. = 0.5 Liter

In the flask, 0.2 mole H2 and 1.3 mole HI are present.

So, molarity of H2 = 0.2/ 0.5 mole/ Lit = 0.4 M

Molarity of HI = 1.3/ 0.5 mole/ Lit = 2.6 M

Equilibrium constant of the reaction is : K = [HI]2 / ([H2 ] * [I2 ] )

So, 0.387 = (2.6)2 / (0.4 * [I2 ] )

Or, [I2 ] = (2.6)2 / (0.4 *0.387) M = 43.67 M

Equilibrium concentration of I2 is 43.67 M.

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---