27. For the given reaction, Qc = [SO3]^2/[SO2]^2.[O2] feeding the concentration
ID: 700274 • Letter: 2
Question
27. For the given reaction,
Qc = [SO3]^2/[SO2]^2.[O2]
feeding the concentration values,
Qc = (10)^2/(0.1)^2.(0.1) = 1.0 x 10^5
So, Qc < Kc, therefore,
d. Qc < Kc, the reaction proceeds from right to left to attain equilibrium
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28. For the given reaction,
from Expt. 1 and 2, [OH-] = constant
when concentration of [ClO2] is doubled, rate quadrupled, so order with respect to [ClO2] = 2
From Expt. 2 and 3, [ClO2] = constant
when concentration of [OH-] is halved, rate also halved, therefore, order with respect to [OH-] is 1
Rate equation thus becomes,
b. rate = k[ClO2]^2.[OH-]
Explanation / Answer
reaction 2SOH) + Ong)290,(g) has the epilibrium constant K.-4.3x toy concentrations are preset 150.]-0.10 M: [SO,1 10. M. IO,1-0.10 M 27 At 700 K, the 250(g)+Oxp) and the following Which of the following is true based on the above? a >K the reaction proceeds from lefi to right to reach equilibrium b>K, the reaction proceeds from right to left to reach equilibriom c. Q.
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