1.If Kp = 845 for the reaction below at 310.0 K, then what is the value of Kc?(R = 0.0821 L・atm/mol・K.) 2 A (g) + B (s) ⇌ 2 C (s) + D (g) 2.A reaction vessel initially contains only 3.5 atm of N₂O₄. When the system reaches equilibrium according to the reaction below, there is 1.7 atm of NO₂ in the vessel. What is Kp for this reaction? 2 NO₂ (g) ⇌ N₂O₄ (g) 3.A solution is made initially with 0.440 M HIO₃ (Kc = 0.170). Once the equilibrium below is established, what

Chemistry: The Molecular Science
5th Edition
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Chapter16: Thermodynamics: Directionality Of Chemical Reactions
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1.If Kp = 845 for the reaction below at 310.0 K, then what is the value of Kc?(R = 0.0821 L・atm/mol・K.) 2 A (g) + B (s) ⇌ 2 C (s) + D (g)

2.A reaction vessel initially contains only 3.5 atm of N₂O₄. When the system reaches equilibrium according to the reaction below, there is 1.7 atm of NO₂ in the vessel. What is Kp for this reaction? 2 NO₂ (g) ⇌ N₂O₄ (g)

3.A solution is made initially with 0.440 M HIO₃ (Kc = 0.170). Once the equilibrium below is established, what is the equilibrium concentration of H⁺ ions? HIO₃ (aq) ⇌ H+ (aq) + IO₃⁻ (aq)

4.An empty steel container is filled with 0.0580 atm of HF. The system is allowed to reach equilibrium. If Kp = 2.76 for the reaction below, what is the equilibrium partial pressure of H₂? 2 HF (g) ⇌ H₂ (g) + F₂ (g)

Expert Solution
Step 1

(1) By help of Following formula we can calculate the value of Kc -

Kp = K(RT)∆n

 

Here , Kp is Equilibrium constant for Gaseous Reaction (845)

Kc is Equilibrium constant for concentration 

R is gas constant = 0.0821 L atm/mol K

T is temperature= 310 K

∆n is number of moles changed in Reaction 2 A (g) + B (s) ⇌ 2 C (s) + D (g) , it is 1-2 = -1 

Then,

845 = Kc (0.0821×310)-1

Kc = 845 × (0.0821 × 310 ) = 21506.095

Kc = 21506.1 

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