K for the reaction 2 HI (g) <-> H2(g) + I2(g) is 0.0184 at a certain temperature

Question

K for the reaction 2 HI (g) <-> H2(g) + I2(g) is 0.0184 at a certain temperature. If 0.100 moles of allthree species were introduced into a 2.50 L container, how manymoles of HI (g) are in the container at equilibrium? A. 0.0400 mol B. 0.0272 mol C. 0.0544 mol D. 0.236 mol E. 0.42 mol 2 HI (g) <-> H2(g) + I2(g) is 0.0184 at a certain temperature. If 0.100 moles of allthree species were introduced into a 2.50 L container, how manymoles of HI (g) are in the container at equilibrium? A. 0.0400 mol B. 0.0272 mol C. 0.0544 mol D. 0.236 mol E. 0.42 mol

Explanation / Answer

D. first you had to find Q to see which way the reaction is going. Q=1 and that is smaller than the K so it is going to the reactant.(if you didnt know how i got that, use the initial concentrationwith regular equilibrium expression for Q) So, 2HI H2 + I2 .04M     .04M .04M ( mols overliters= molarity what we use) +2x        -x      -x      (regular ice box) .04+2X   .04M-x .04-x     k=.0184 [.04-x][.04-x]/ [.04+2x]^2 = .0184 = [.04-x]^2 / [.04+2x]^2 = .0184      squareroot both sides and you get [.04-x]/[ .04+2x]= .1356 [.04-x]= .00542+ .2712x .04-.00542=1.2712x .03458=1.2712x .02720=x     this is the value of x not theconcentration of HI, you plug it back in! so .04+2x   where x=.02720 [HI]= .0944M but you want it in mols so you divide by 2.50L cuz Mis mols/L (.0944mols/L)(2.50L)=.236 mol hope that helps! PLEASE RATE!!! thx

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