The following concentrations were obtained for the formation of NH3 from N2 and H2 at equilibrium at 500 K. [N2] = 1.5 x 10-2 M. [H2] = 3.0 x 10-2 M and [NH+3] = 1.2 x 10-2 M. The equilibrium constant is

Question

The following concentrations were obtained for the formation of NH3 from N2 and H2 at equilibrium at 500 K. [N2] = 1.5 x 10-2 M. [H2] = 3.0 x 10-2 M and [NH+3] = 1.2 x 10-2 M. The equilibrium constant is (A) 4.06 × 10-2  (B) 3.55 × 102  (C) 3 × 104  (D) 3.5 × 10-7

Tardigrade Admin · Accepted Answer

The equilibrium constant for the reaction N2g + 3H 2g leftharpoons 2NH3g can be written as, kc =([NH3h]2/[N2g][H2g]3 ) = ((1.2×10-2)2/(1.5×10-2)(3.0×10-2)3)=3.55×102

Q. The following concentrations were obtained for the formation of $N H_{3}$ from $N_{2}$ and $H_{2}$ at equilibrium at $500 K . [N_{2}] = 1.5 x 1 0^{- 2} M . [H_{2}] = 3.0 x 1 0^{- 2} M$ and $[N H + 3] = 1.2 x 1 0^{- 2} M .$ The equilibrium constant is

$4.06 \times 1 0^{- 2}$

$3.55 \times 1 0^{2}$

$3 \times 1 0^{4}$

$3.5 \times 1 0^{- 7}$

The equilibrium constant for the reaction
$N_{2} g + 3 H_{2} g ⇌ 2 N H_{3} g$ can be written as,
$k_{c} = \frac{[ N H _{3 h} ] ^{2}}{[ N _{2 g} ] [ H _{2 g} ] ^{3}}$
$= \frac{( 1.2 \times 1 0 ^{- 2} ) ^{2}}{( 1.5 \times 1 0 ^{- 2} ) ( 3.0 \times 1 0 ^{- 2} ) ^{3}} = 3.55 \times 1 0^{2}$

Q. The following concentrations were obtained for the formation of NH3​ from N2​ and H2​ at equilibrium at 500K.[N2​]=1.5x10−2M.[H2​]=3.0x10−2M and [NH+3]=1.2x10−2M. The equilibrium constant is

4.06×10−2

3.55×102

3×104

3.5×10−7