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Question 2 CH13 1) For the reaction 2A(g)+2B(g)⇌C(g) Kc = 68.4 at a temperature of 311...

Question 2 CH13

1) For the reaction 2A(g)+2B(g)⇌C(g) Kc = 68.4 at a temperature of 311 ∘C . Calculate the value of Kp. Express your answer numerically.

For the reaction

X(g)+3Y(g)⇌2Z(g)

Kp = 1.76×10⁻² at a temperature of 255 ∘C .

Calculate the value of Kc.

Solutions

Expert Solution

2A(g)+2B(g)⇌C(g)

Kp = Kc (RT)ⁿ

n = total no of moles of products in gases - total no of moles of reactants in gases

= 1-(2+2) = -3

T = 311C⁰ = 311+273 = 584K

Kp = Kc (RT)ⁿ

= 68.4(0.0821*584)^-3

= 68.4(47.9464)^-3

= 68.4/11022193035224 = 6.2*10^-12

X(g)+3Y(g)⇌2Z(g)

Kp = Kc (RT)ⁿ

n = total no of moles of products in gases - total no of moles of reactants in gases

= 2-(1+3) = -2

T = 255C⁰ = 255+273 = 528K

R = 0.0821L-atm/mole-K

Kp = Kc (RT)ⁿ

1.76*10^-2 = Kc(0.0821*528)^-2

1.76*10^-2 = Kc(43.3488)^-2

1.76*10^-2 = Kc/1879.1184

KC = 1.76*10^-2 *1879.1184

= 33.0724>>> answer

queen_honey_blossom answered 1 year ago

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