Answer0.05 Question 11 Consider the formation of hydrogen fluoide Not changed si

Question

Answer0.05 Question 11 Consider the formation of hydrogen fluoide Not changed since last attempt Marked out of 1.00 H2g)+F2(9)--2HF(g) If a 1.5 L nickel reaction container (glass cannot be used equlibrium constant, Kp, is 7.8 x 1014 (Hint, this is a very large number, what does that imply?) Calculate the molar concentration of HF at equilibrium. A further hint is provided after the first attempt in the feedback because it reacts with HF) filled with 0 0086 M H2 is connected to a 3.2 L container filed with 0.030 M F2 The P Flag question Answer. 0.040 Question 12 BONUS Suppose a 1 00 L nickel reaction container filled with 0 0080 M H2 is connected to a 4 00 equilibrium L container thilled with 0.029 M F2. Calculate the molar concentration of H2 at Not yet answered Marked out of 1.00 A hint is provided after the first attempt in the feedback v Flag question Answer. Question 13 Sulfur dioxide reacts with chlorine at 227 °C Not yet answered Marked out of 1.00 ? Flag question Kp for lhis reaction is 5 1 × 1o? atm. Initially, 1 00 g each of SO2 and Cl2 are placed in a 1 00 L reaction vessel Ater 15 minutes, You will determine if the system has reached equilibrium. First, what is Ke (in UmOI)? (A g is 106 g )

Explanation / Answer

initial number of moles of H2 = 0.0086 * 1.5 = 0.0129

initial number of moles of F2 = 0.03*3.2 = 0.096

Total volume of the container = 1.5 + 3.2 = 3.7 L

             H2(g)      +     F2(g)   <-----> 2HF(g)

initially 0.0129/3.7      0.096/3.7            0

           0.003486          0.02594            0

at equili (0.003486 -x)    (0.02594 - x)       2x

Kc = [HF]^2/[H2][F2]

Kc = (2x)^2/((0.003486 -x)*(0.02594 - x))

Kc = Kp*(RT)^dn

dn = 2-2 = 0 (gas moles on product side - gas moles on reactant side)

Kc = Kp

7.8*10^14 = (2x)^2/((0.003486 -x)*(0.02594 - x))

by solving x = 0.003486

at equlibrium [H2] = 0

[F2] = 0.02594 - 0.003486 = 0.02245

[HF] = 2*0.003486 = 0.006972 M

initial number of moles of H2 = 0.008 * 1 = 0.008

initial number of moles of F2 = 0.029*4 = 0.116

Total volume of the container = 1 + 4 = 5 L

             H2(g)    +     F2(g)   <-----> 2HF(g)

initially 0.008/5        0.116/5            0

           0.0016          0.0232            0

at equili (0.0016 -x)    (0.0232 - x)       2x

Kc = [HF]^2/[H2][F2]

Kc = (2x)^2/((0.0016 -x)*(0.0232 - x))

Kc = Kp*(RT)^dn

dn = 2-2 = 0 (gas moles on product side - gas moles on reactant side)

Kc = Kp

7.8*10^14 = (2x)^2/((0.0016 -x)*(0.0232 - x))

x = 0.0016

at equlibrium [H2] = 0

[F2] = 0.0232 - 0.0016 = 0.0216 M

[HF] = 2*0.0016 = 0.0032 M

            SO2        +    Cl2 <------> So2Cl2

initially   (1/64*1/1)   (1/71*1/1)      0

             0.01562      0.01408        0

at equili    (0.01562-x) (0.01408-x)    x = 45.5 microg/L

Kc = [SO2Cl2]/[SO2][Cl2]

Kc = (45.5*10^-6)/((0.01562-(45.5*10^-6))*(0.01408-(45.5*10^-6)))

Kc = 0.2082

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