In: Chemistry
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.
2)
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.
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
2
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