18. 2HI(g) Ho(g)+1(g) Kc-0.0180 A 5.00 L reaction vessel is filled with 4.00 mol

Question

18. 2HI(g) Ho(g)+1(g) Kc-0.0180 A 5.00 L reaction vessel is filled with 4.00 moles of HI. How many moles of H2 will be present at equilibrium? A. 0.0139 mol B. 0.134 mol C. 0.0288 mol D. 0.423 mol 19. A student ran the following reaction in the laboratory at 311 K: When she introduced 0.0429 moles of CH4 and 0.0669 moles of CCl4 into a 5.00 L container, she found the equilibrium concentration of CH Cl2 to be 0.00997 M. Calculate the Kc that she obtained from her data. A. 0.866 B. 3.29 C. 0.347 D. 0.0346 e more page

Explanation / Answer

2HI(g) -------------> H2(g) + I2(g)
I   4                      0       0
C   -2x                     +x     +x
E   4-2x                    +x      +x
     Kc   = [H2][I2]/[HI]^2
     0.018 = x*x/(4-2x)^2
     0.018*(4-2x)^2 = x^2
         x = 0.423
     no of moles H2 = x = 0.423mole
D. 0.423mol >>>>>answer
19.
[CH4] = no of moles/volume = 0.0429/5   = 0.00858M
[CCl4] = no of moles/volume = 0.0669/5   = 0.01338M

     CH4(g)    + CCl4(g) -----------> 2CH2Cl2
I    0.00858     0.01338               0
C    -0.004985   -0.004985            2*0.004985
E     0.003595    0.008395            0.00997

        Kc   = [CH2Cl2]^2/[CH4][CCl4]
             = (0.00997)^2/(0.003595)*(0.008395) = 3.29
B.3.29

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