Please show how you got the answers. Thank you so much! 21. The gas-phase decomp

Question

Please show how you got the answers. Thank you so much! 21. The gas-phase decomposition of dinitrogen pentoxide leads to products N0sg products At 332 K the rate constant for this reaction is 2.5 x 10 s and at 375 K the rate constant is 1.5 x 10 s. What is the Ea (J/mol) for this reaction. A. 2.1 x 10 B. 1.0 x 10 C. 3.5 x 106 D. 9.8 x 104 22. What will happen to the equilibrium in the reaction 2A,g) «OOLs) C(g) Kc = 1.25 at 300 K + , if a the volume is reduced. A. The reaction is forced to the right. B. The reaction is forced to the left. C. No change D. K is increased. E. Kc is decreased. 23. ½Ogg) at 600 Calculate KP for K, using the following data: H,og)+ H,02(g) « Kr = 2.3 x 10" K, = 2.35 x 10.19 H,Ogg) “Hap +Ogg) o A. 4.4 x1043 B. 9.8 x 1024 C. 1.2 x 104 D. 1.0 x 1013 E. 2.6 x 1031

Explanation / Answer

21) Arrhenius equation

ln(K2/k1) = Ea/R[1/T1 - 1/T2]

k1 = 2.5*10^-3 s-1 , T1 = 332 k

k2 = 1.5*10^-1 s-1 , T2 = 375 k

Ea = ? kj/mol , R = 8.314*10^-3 j.k-1.mol-1

ln(0.15/(2.5*10^-3)) = ((x*10^3)/(8.314))((1/332)-(1/375)

Ea = activation energy = 98.56 kj/mol

answer: D.9.8*10^4j/mol

22) volume decreases,pressure increases.so that, equilibrium shift towards decreasing the pressure (less no of mol side).

answer: A. The reaction is forced to the right.

23) from the given data,

take reverse of eq1   , K1 = (1/2.3*10^-6)

   eq2 ,                   K2 = 2.35*10^-19

equilibrium constant(K) = k1*k2

                         = ((1/(2.3*10^-6))*(2.35*10^-19)

                         = 1.0*10^013
answer: D

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