The equilibrium constant, K, for the following reaction is 1.20×102 at 500 K. PCI5(g)=PCI3(g) + Cl(g) Calculate the equilibrium concentrations of reactant and products when 0.331 moles of PCl<(g) are introduced into a 1.00 L vessel at 500 K. [PCI3] = M %3D [PCI3] = M [Cl,] M
The equilibrium constant, K, for the following reaction is 1.20×102 at 500 K. PCI5(g)=PCI3(g) + Cl(g) Calculate the equilibrium concentrations of reactant and products when 0.331 moles of PCl<(g) are introduced into a 1.00 L vessel at 500 K. [PCI3] = M %3D [PCI3] = M [Cl,] M
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Chapter1: Chemical Foundations
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I dont know how to do these two questions on Equilibrium concentration
![The equilibrium constant, \( K_c \), for the following reaction is \( 1.20 \times 10^{-2} \) at 500 K.
\[ \text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g) \]
Calculate the equilibrium concentrations of reactant and products when 0.331 moles of \( \text{PCl}_5(g) \) are introduced into a 1.00 L vessel at 500 K.
\[
[\text{PCl}_5] = \_\_\_\_\_\_\_\_\_\_ \text{ M}
\]
\[
[\text{PCl}_3] = \_\_\_\_\_\_\_\_\_\_ \text{ M}
\]
\[
[\text{Cl}_2] = \_\_\_\_\_\_\_\_\_\_ \text{ M}
\]
Buttons for submitting the answer and retrying the entire group of questions are available. There are 9 more group attempts remaining.
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Transcribed Image Text:The equilibrium constant, \( K_c \), for the following reaction is \( 1.20 \times 10^{-2} \) at 500 K.
\[ \text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g) \]
Calculate the equilibrium concentrations of reactant and products when 0.331 moles of \( \text{PCl}_5(g) \) are introduced into a 1.00 L vessel at 500 K.
\[
[\text{PCl}_5] = \_\_\_\_\_\_\_\_\_\_ \text{ M}
\]
\[
[\text{PCl}_3] = \_\_\_\_\_\_\_\_\_\_ \text{ M}
\]
\[
[\text{Cl}_2] = \_\_\_\_\_\_\_\_\_\_ \text{ M}
\]
Buttons for submitting the answer and retrying the entire group of questions are available. There are 9 more group attempts remaining.
Navigation buttons for "Previous" and "Next" are also present.
This exercise is from a Cengage Learning platform, with technical support options available.
![**Equilibrium Concentrations Calculation**
The equilibrium constant, \(K_c\), for the following reaction is \(1.80 \times 10^{-2}\) at 698 K.
\[ 2HI(g) \rightleftharpoons H_2(g) + I_2(g) \]
**Objective:**
Calculate the equilibrium concentrations of reactant and products when 0.370 moles of HI are introduced into a 1.00 L vessel at 698 K.
**Equilibrium Concentration Boxes:**
- \([HI] = \_\_\_\_\_\_\_\_ \) M
- \([H_2] = \_\_\_\_\_\_\_\_ \) M
- \([I_2] = \_\_\_\_\_\_\_\_ \) M
**Instructions:**
Enter the calculated concentrations of HI, H₂, and I₂ in the provided fields. Use the equilibrium constant to set up and solve the equilibrium expression.
Submit your answers by clicking "Submit Answer." You can retry the problem or move to the next one using "Retry Entire Group." You have 9 more group attempts remaining.
**Notes:**
- \(K_c\) is provided for calculations to assess shifts in equilibrium.
- Ensure that calculations consider the initial concentration reduction and any stoichiometric relationships.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F00585539-79e1-42a7-8a75-95bc6de8d9d5%2F2a104dba-fe81-4433-927b-6a7226bb44b6%2Fq51dlen_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Equilibrium Concentrations Calculation**
The equilibrium constant, \(K_c\), for the following reaction is \(1.80 \times 10^{-2}\) at 698 K.
\[ 2HI(g) \rightleftharpoons H_2(g) + I_2(g) \]
**Objective:**
Calculate the equilibrium concentrations of reactant and products when 0.370 moles of HI are introduced into a 1.00 L vessel at 698 K.
**Equilibrium Concentration Boxes:**
- \([HI] = \_\_\_\_\_\_\_\_ \) M
- \([H_2] = \_\_\_\_\_\_\_\_ \) M
- \([I_2] = \_\_\_\_\_\_\_\_ \) M
**Instructions:**
Enter the calculated concentrations of HI, H₂, and I₂ in the provided fields. Use the equilibrium constant to set up and solve the equilibrium expression.
Submit your answers by clicking "Submit Answer." You can retry the problem or move to the next one using "Retry Entire Group." You have 9 more group attempts remaining.
**Notes:**
- \(K_c\) is provided for calculations to assess shifts in equilibrium.
- Ensure that calculations consider the initial concentration reduction and any stoichiometric relationships.
Expert Solution

Step 1
Since you have posted multiple questions, we are entitled to answer the first only.
The equilibrium reaction given is,
Given: Kc = 0.0120
Initial moles of PCl5 = 0.331 mol.
And the volume of flask = 1.00 L
Step by step
Solved in 2 steps

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