Consider the following reaction: 3N_2H_4(g) + 4ClF_3(g) doubleheadarrow 12HF(g)

Question

Consider the following reaction: 3N_2H_4(g) + 4ClF_3(g) doubleheadarrow 12HF(g) + 3N_2(g) + 2Cl_2(g) A mixture, initially consisting of 0.880 M N_2H_4(g) and 0.880 M CIF_3(g), reacts at a certain temperature. At equilibrium, the concentration of N_2(g) is 0.525 M. Calculate the concentration of HF(g) at equilibrium. 0.525 M 0.700 M 2.10 M 0.705 M 0.175 M Consider the following reaction: 2SO_2(g) + O_2(g) doubleheadarrow 2SO_3(g) If the value of K_c for this reaction is 13 at 900 K, calculate the value of at 900 K. 0 .18 960 13 0.0024 0.077 What is the relationship between K_p and K_c for the reaction below? At 600degree C, the equilibrium constant for the reaction 2HgO(s) doubleheadarrow 2Hg(1) + O_2(g) is 2.8. Calculate the equilibrium constant for the reaction

Explanation / Answer

Molar ratio of N2H4: ClF3= 3 :4 = 1 :1.33

Given ratio of N2H4: ClF3= 1:1 .Limiting reactant is ClF3

Let x= concentration ClF3 that decomposed , then (3/4)*x M of N2H4 also decompses

ClF3 at equilibrium = 0.880-x N2H4 at equilibrium= 0.88-0.75x

HF fomed= 12x and Cl2= 2x and N2= 3x

But 3x= 0.525, x= 0.525/3=0.175M HF at equilibrium= 12*0.175=2.1M ( C is the correct answer)

Kp =Kc(RT)deln

R= 0.0821 L.atm/mole.Kand T= 900K

Deln= moles of products- mole sof reactans = 2-(2+1)=-1

Given Kc= 13

Kp = 13*(0.0821*900 )-1=0.1759 ( A is the correct answer)

Kp =Kc(RT)deln

Deln= moles of products- mole sof reactans = 2+1-2= 1

Kp= KCRT( D is the correct answer)

K= [O2] [Hg]2/ [HgO]2= 2.8

Sqrt(K)= [O2]1/2 [Hg] /[HgO]=sqrt(2.8)=1.673

For the reaction 1/O2+Hg-à HgO

K1= [HgO] / [Hg] [O2]1/2 =1/sqrt(K)= 1/1.673=0.5976 (A is the correct answer)

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