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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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To determine whether the reaction mixture is at equilibrium, we need to calculate the reaction quotient \( Q_c \) and compare it to the equilibrium constant \( K_c \). ### Step 1: Write the balanced chemical equation The balanced chemical equation for the reaction is: \[ N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \] ### Step 2: Calculate the concentrations of the reactants and products To find the concentrations, we use the formula: ...
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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 ?

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CENGAGE CHEMISTRY-NCERT BASED EXERCISE-Chemical Equilibrium
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