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A mixture of 1.57 mol of N(2), 1.92 mol ...

A mixture of `1.57 mol` of `N_(2), 1.92 mol` of `H_(2)` and `8.13 mol` of `NH_(3)` is introduced into a `20 L` reaction vessel at `500 K`. At this temperature, the equilibrium constant `K_(c )` for the reaction `N_(2)(g)+3H_(2)(g) hArr 2NH_(3)(g)` is `1.7xx10^(2)`. Is the reaction mixture at equilibrium? If not, what is the direction of the net reaction?

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A mixture of 2 moles of N_2,2.0 moles of H_2 and 10.0 moles of NH_3 is introduced into a 20.0 L reaction vessel at 500 K . At this temperature, equilibrium constant K_c is 1.7 xx 10^2 , for the reaction N_2(g) +3H_2(g) hArr 2NH_3(g) (i) is the reaction mixture at equilibrium ? (ii) if not, what is the direction of the reaction ?

A mixture of 1.57 mol of N_(2) ,1.92 mol of H_(2) and 8.17 mol of NH_(3) is introduced in a 20 L reaction vessel at 500 K. At this temperature, the equilibrium constant K_(c) for the reaction is 1.7 xx 10^(-2) . Is the reaction at equilibrium? The reaction is N_(2)(g)+3H_(2)(g) rarr 2NH_(3)(g)

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For the reaction, N_(2)(g)+3H_(2)(g) hArr 2NH_(3)(g) , the units of K_(p) are …………

The equilibrium constant K_(p) , for the reaction N_(2)(g)+3H_(2)(g) hArr 2NH_(3)(g) is 1.6xx10^(-4) at 400^(@)C . What will be the equilibrium constant at 500^(@)C if the heat of reaction in this temperature range is -25.14 kcal?

At 773 K, the equilibrium constant K_(c) for the reaction, N_(2) (g) + 3 H_(2) (g) hArr 2 NH_(3) (g)" is " 6.02 xx 10^(-2)L^(2) mol^(-2). Calculate the value of K_(p) at the same temperature.

At 800 K, the equilibrium constant for the reaction, N_2(g) + 3H_2(g) hArr 2NH_3(g) is 6.05xx10^(-2)L^2 "mol"^(-2) . Calculate K_p for the reaction at the same temperature.

RESONANCE-CHEMICAL EQUILIBRIUM-Exercise-1 (Part-1)
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  14. In a container H(2)O(g),CO(g) "and" H(2)(g) are present in the molar r...

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