aleks.com/alekscgi/x/Isl.exe/1o u-lgNslkr7j8P3jH-Uxk3WxUPZu2ctUcqMoMncl_czxr-fFx

Question

aleks.com/alekscgi/x/Isl.exe/1o u-lgNslkr7j8P3jH-Uxk3WxUPZu2ctUcqMoMncl_czxr-fFxXXD8-fh1WHbIRMDSAQhdNxjJ6nmlZjC-sPfSSCnc40C6q0GPmoPVC -O KINETICS AND EQUIUBRIUM Calculating equilibrium composition from an equilibrium constant Suppose a 500. mL flask is filled with 0.80 mol of H2 and 1.7 mol of HI. The following reaction becomes possible: H2(8)+1)2HI(g) The equilibrium constant K for this reaction is 7.84 at the temperature of the flask. Calculate the equilibrium molarity of I2. Round your answer to two decimal places.

Explanation / Answer

H2   + I2 <--------> 2HI

mol mol mol

I 0.80 1.70 0

C -x -x +x

E (0.80 -x) (1.70 -x) x

Equilibrium Constant, Kc = [HI]2/ [H2][I2]

7.84 = x2/ (0.80-x)(1.70-x)

Solving for x, we get x= 2.14 or x=0.73. Logically possible x value should be 0.73

So moles of HI = 0.73,

Equilibrium concentrations:

HI, [HI] = 0.73 moles/0.500 L = 1.46 M

I2,[I2] = (1.70 -0.73) moles /0.500L = 1.94 M

H2,[H2] = (0.80 -0.73) moles/0.500L = 0.14 M

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.