for the decomposition of 1 mol of nh3 into N2 and H2 Kc=2x10^-3 if the equilibrium concentration of n2 is 0.09 M and h2 is 0.04 M then the equilibrium concentration of nh3 is (a) 1.2 M (b) 2.88x10^-3 M (c) 5.36x10^-2 M (d) 1.34 M

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Adarsh · Answer

kc=(N2).(H2)^3/(NH3)^2=>0.09 x 0.04 x 0.04 x 0.04 /(NH3)^2= 2 x 10^-3=> (NH3) = 5.36 x 10^-2 hence c is the correct option.

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for the decomposition of 1 mol of nh3 into N2 and H2 Kc=2x10^-3 if the equilibrium concentration of n2 is 0.09 M and h2 is 0.04 M then the equilibrium concentration of nh3 is (a) 1.2 M (b) 2.88x10^-3 M (c) 5.36x10^-2 M (d) 1.34 M

for the decomposition of 1 mol of nh3 into N2 and H2 Kc=2x10^-3 if the equilibrium concentration of n2 is 0.09 M and h2 is 0.04 M then the equilibrium concentration of nh3 is (a) 1.2 M (b) 2.88x10^-3 M (c) 5.36x10^-2 M (d) 1.34 M

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for the decomposition of 1 mol of nh3 into N2 and H2 Kc=2x10^-3 if the equilibrium concentration of n2 is 0.09 M and h2 is 0.04 M then the equilibrium concentration of nh3 is (a) 1.2 M (b) 2.88x10^-3 M (c) 5.36x10^-2 M (d) 1.34 M

for the decomposition of 1 mol of nh3 into N2 and H2 Kc=2x10^-3 if the equilibrium concentration of n2 is 0.09 M and h2 is 0.04 M then the equilibrium concentration of nh3 is (a) 1.2 M (b) 2.88x10^-3 M (c) 5.36x10^-2 M (d) 1.34 M

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