when 8.4 mol of PCl5 (g) is mixed with 1.8 mol of PCl3 (g) and allowed to come to equilibrium in a 10 dm3container the amount of PCl5 (g) at equilibrium is 7.2 mol. Kc = (b) Explain the effect of the following changes below on the value of Kc: Increasing temperature Lowering the concentration
when 8.4 mol of PCl5 (g) is mixed with 1.8 mol of PCl3 (g) and allowed to come to equilibrium in a 10 dm3container the amount of PCl5 (g) at equilibrium is 7.2 mol. Kc = (b) Explain the effect of the following changes below on the value of Kc: Increasing temperature Lowering the concentration
Chemistry
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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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(a) Calculate the value of Kc for the reaction:
PCl5 (g) PCl3 (g) + Cl2 (g) ΔH = Positive
Given that when 8.4 mol of PCl5 (g) is mixed with 1.8 mol of PCl3 (g) and allowed to come to equilibrium in a 10 dm3container the amount of PCl5 (g) at equilibrium is 7.2 mol.
Kc =
(b) Explain the effect of the following changes below on the value of Kc:
- Increasing temperature
- Lowering the concentration of chlorine (Cl2)
- Addition of a catalyst
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Follow-up Question
i believe your ans for Kc is incorrect
with Kc = 0.05
![Examples of cal. Kc.
Calculate the value of Kc for the reacher.
PCLSS
= P C3(g) + Cl₂(g).
Given that when 8.4 mol of PCs (s) is mixed
with 1-8 mor of PCL3(g) & allowed to come
to equilibrium in a 10dm³ container the
amount of PC`s(s) at equilibrum is 7.2 mol.
PCL5
- PCI 3
ndes initially 8.4
84-102
7-2
nores@eqmail.
Full conc in
mol/dm³
7-2
10.
13.29
+ Cl₂.
1.8.
(²+1.8) 1.2.
3.0/10
1-2
10.
= 0-72
0.30.
012.
kc = [PC13 ] [C1₂] = [03][012]=K=00
[PCL5]
En = takes in
Ex = gives
+
out.
ma](https://content.bartleby.com/qna-images/question/824ec50a-9f05-4b47-b4de-079791e47d11/68c5e66e-4876-49a0-93c8-3ba008313654/47o8smj_thumbnail.jpeg)
Transcribed Image Text:Examples of cal. Kc.
Calculate the value of Kc for the reacher.
PCLSS
= P C3(g) + Cl₂(g).
Given that when 8.4 mol of PCs (s) is mixed
with 1-8 mor of PCL3(g) & allowed to come
to equilibrium in a 10dm³ container the
amount of PC`s(s) at equilibrum is 7.2 mol.
PCL5
- PCI 3
ndes initially 8.4
84-102
7-2
nores@eqmail.
Full conc in
mol/dm³
7-2
10.
13.29
+ Cl₂.
1.8.
(²+1.8) 1.2.
3.0/10
1-2
10.
= 0-72
0.30.
012.
kc = [PC13 ] [C1₂] = [03][012]=K=00
[PCL5]
En = takes in
Ex = gives
+
out.
ma
Solution
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