Answer:
Kc = 3.72 × 10⁶
Explanation:
Let's consider the following reaction:
NH₄HS(g) ⇄ NH₃(g) + H₂S(g)
At equilibrium, we have the following concentrations:
[NH₄HS] = 0.196 M (assuming a 1 L flask)
[NH₃] = 9.56 × 10² M
[H₂S] = 7.62 × 10² M
We can replace this data in the Kc expression.
Answer : The % of (+) limonene isomer = 79%
The % of (-) limonene isomer = 0%
The % of enantiomeric excess = 58%
Explanation : Enantiomeric excess (ee) is the measurement of purity used for chiral substances.
Given,
% of pure limonene enantiomer = The % of (+) limonene isomer = 79%
Therefore, The % of (-) limonene isomer = 0%
Formula used :
Where, ee → enantiomeric excess
Now, put all the values in above formula, we get the value of enantiomeric excess (ee).
= 58%