Question

Part A For the reaction 3A(g)+3B(g)⇌C(g) Kc = 35.0 at a temperature of 347 ∘C . Calculate the value of Kp. Express your answer numerically. Kp = SubmitHintsMy AnswersGive UpReview Part

Part B For the reaction X(g)+3Y(g)⇌3Z(g) Kp = 2.30×10−2 at a temperature of 271 ∘C . Calculate the value of Kc. Express your answer numerically. Kc = SubmitHintsMy AnswersGive UpReview Part

Answer 1

Part A Solution Given Kc- 35.0 T-347°C- 620.15K R = 0.08314 L bar Kl mol-I R-0.08206 L atm Kmol 3A(g)+3B(Cg) Where An(g-¢ moles of gaseous product)-(E moles of gaseous reactant) Kp-35.0x(0.08314L barK1 morx620.15K)-9.606 x10 Kp 35.0x (0.08206 L atmK mol1x620.15 K)-1.025 x10 Ko-9. 60бы 10-1 (in tems of bar) Kp-1.025x10(in terms of atm)

Part B Solution Given Kp-2.30x10-2 T 271 °C 544.15 K Where An(g)- (2 moles of gaseous product) - (2 moles of gaseous reactant) insert An(g)--1 Kp-Kc(RT) weget, Kc- Kp(RT) Kc_ 2.30 10-3x (0.083 14 L bar K-1 mol#x 544.15 K)-1.041 Kc 230x10x (008206 L atm K mol x 544.15K)-1.027 c1.041] (Kp in tems of bar) Kc 1.027 (Kpin terms of atm)