Hi, Need help with the following Q PartA & PartB open image in new tab for bette

Question

Hi,

Need help with the following Q PartA & PartB

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Part A

For the reaction

2A(g)+2B(g)C(g)

Kc = 32.2 at a temperature of 351 C .

Calculate the value of Kp.

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Part B

For the reaction

X(g)+2Y(g)3Z(g)

Kp = 2.17×102 at a temperature of 131 C .

Calculate the value of Kc.

CH13 H ± Pressure-Based versus Concentration-Based Equilibrium Constants ± Pressure-Based versus Concentration-Based Equilibrium Constants Part A The equilibrium constant, Kis calculated using molar concentrations. For gaseous reactions another form of the equilibrlum constant, Kp is calculated from partial pressures instead of concentrations. These two equilibrium constants are related by the equation For the reaction 2A(g) + 2B(g) C(g) Ke 32.2 at a temperature of 351 "C Calculate the value of Kp Express your answer numerically. where R = 0.08206 L-atm/(K·mol). T is the absolute temperature, and n is the change in the number of moles of gas (sum moles products - sum moles reactants). For axample, consider the reaction Hints N2(g) +3H2(g) 2NH3 (g) for which n=2 (1+3)= 2 Submit My AnsweES Give Up Incorrect; Try Again; 3 attempts remaining Part B For the reaction Kp 2.17-10-2 at a temperature of 131 C Calculate the value of K Express your answer numerically. Hints Submit My Answers Give Up

Explanation / Answer

A)
T= 351.0 oC
= (351.0+273) K
= 624 K
n = number of gaseous molecule in product - number of gaseous molecule in reactant
n = -3
Kp= Kc (RT)^ n
Kp = 32.2*(0.0821*624.0)^(-3)
Kp = 2.395*10^-4
Answer: 2.395*10^-4

B)
T= 131.0 oC
= (131.0+273) K
= 404 K
n = number of gaseous molecule in product - number of gaseous molecule in reactant
n = 0
Kp= Kc (RT)^n
0.0217 = Kc *(0.0821*404.0)^(0)
Kc = 2.17*10^-2

Answer: 2.17*10^-2

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