The decomposition of NH3 on platinum surface is zero order reaction. What are the rates of production of N2 and H2 if k=2.5× 10- 4molL- 1sec- 1 ? 2 NH 3( g ) arrow N 2( g )+3 H 2( g )

Question

The decomposition of NH3 on platinum surface is zero order reaction. What are the rates of production of N2 and H2 if k=2.5× 10- 4molL- 1sec- 1 ? 2 NH 3( g ) arrow N 2( g )+3 H 2( g ) (A) 1.25× 10- 4 Ms- 1,3.75× 10- 4 Ms- 1 (B) 3.00× 10- 4 Ms- 1,7.50× 10- 4 Ms- 1 (C) 2.50× 10- 4 Ms- 1, 1.25× 10- 4 Ms- 1 (D) 3.75× 10- 4 Ms- 1,2.50× 10- 4 Ms- 1

Tardigrade Admin · Accepted Answer

2 NH 3( g ) arrow N 2( g )+3 H 2( g ) Rate = k[NH3]o = k = 2.5 Ã 10-4 M s-1. (1/2) (- textd/ textdt) [ textNH3] = (1/1) ( textd/ textdt) [ textN text2] = (1/3) ( textd/ textdt) [ textH text2] ( d / d â¡ t) [ textN2] = (2 ⋅ 5 × 1 0- 4/2) = 1 ⋅ 2 5 × 1 0- 4 textM s- 1 ( d / d â¡ t) [ H â¡2] = (3/2) × 2 ⋅ 5 × 1 0- 4 = 3 ⋅ 7 5 × 1 0- 4 textM s- 1

Q. The decomposition of $N H_{3}$ on platinum surface is zero order reaction. What are the rates of production of $N_{2}$ and $H_{2}$ if $k = 2.5 \times 1 0^{- 4} m o l L^{- 1} se c^{- 1}$ ?
$2 N H_{3} (g) \to N_{2} (g) + 3 H_{2} (g)$

$1.25 \times 1 0^{- 4} M s^{- 1}, 3.75 \times 1 0^{- 4} M s^{- 1}$

$3.00 \times 1 0^{- 4} M s^{- 1}, 7.50 \times 1 0^{- 4} M s^{- 1}$

$2.50 \times 1 0^{- 4} M s^{- 1}, 1.25 \times 1 0^{- 4} M s^{- 1}$

$3.75 \times 1 0^{- 4} M s^{- 1}, 2.50 \times 1 0^{- 4} M s^{- 1}$

Q. The decomposition of NH3​ on platinum surface is zero order reaction. What are the rates of production of N2​ and H2​ if k=2.5×10−4molL−1sec−1 ? 2NH3​(g)→N2​(g)+3H2​(g)

1.25×10−4Ms−1,3.75×10−4Ms−1

3.00×10−4Ms−1,7.50×10−4Ms−1

2.50×10−4Ms−1,1.25×10−4Ms−1

3.75×10−4Ms−1,2.50×10−4Ms−1

2NH3​(g)→N2​(g)+3H2​(g) Rate = k[NH3]o = k = 2.5 × 10-4 M s-1. 21​dt−d​[NH3​]=11​{dtd​[N2​]}=31​{dtd​[H2​]} d⁡td​[N2​]=22⋅5×10−4​=1⋅25×10−4M s−1 d⁡td​[H⁡2​]=23​×2⋅5×10−4=3⋅75×10−4M s−1

Q. The decomposition of $N H_{3}$ on platinum surface is zero order reaction. What are the rates of production of $N_{2}$ and $H_{2}$ if $k = 2.5 \times 1 0^{- 4} m o l L^{- 1} se c^{- 1}$ ?
$2 N H_{3} (g) \to N_{2} (g) + 3 H_{2} (g)$

$1.25 \times 1 0^{- 4} M s^{- 1}, 3.75 \times 1 0^{- 4} M s^{- 1}$

$3.00 \times 1 0^{- 4} M s^{- 1}, 7.50 \times 1 0^{- 4} M s^{- 1}$

$2.50 \times 1 0^{- 4} M s^{- 1}, 1.25 \times 1 0^{- 4} M s^{- 1}$

$3.75 \times 1 0^{- 4} M s^{- 1}, 2.50 \times 1 0^{- 4} M s^{- 1}$