The decomposition of NH3 on platinum surface is zero order reaction. The rates of production of N2 and H2 are respectively (k = 2.5 × 10-4 mol L-1 s-1)

Question

The decomposition of NH3 on platinum surface is zero order reaction. The rates of production of N2 and H2 are respectively (k = 2.5 × 10-4 mol L-1 s-1) (A) 7.5 × 10-4 mol L-1 s-1 ;2.5 × 10-4 mol L-1s-1 (B) 2.5 × 10-4 mol L-1 s-1 ;7.5 × 10-4 mol L-1s-1 (C) 2.5 × 10-4 mol L-1 s-1 ;2.5 × 10-4 mol L-1s-1 (D) 2.5 × 10-4 mol L-1 s-1 ;5.0 × 10-4 mol L-1s-1

Tardigrade Admin · Accepted Answer

Given reaction is 2NH3 xrightarrowpt N2 + 3 H2 For zero order reaction, rate = k ∴ -(1/2)(d[NH]3/dt)= + (d[N2]/dt) =+ (1/3)(d[H2]/dt) = 2.5×10-4 mol L-1 s-1 Rate of production of N2 =(d[N2]/dt) =2.5×10-4 mol L-1s-1 Rate of production of H2 =(d[H2]/dt )=3×(2.5×10-4) =7.5×10-4 mol L-1 s-1

Q. The decomposition of $N H_{3}$ on platinum surface is zero order reaction. The rates of production of $N_{2}$ and $H_{2}$ are respectively $(k = 2.5 \times 1 0^{- 4} m o l L^{- 1} s^{- 1})$

$7.5 \times 1 0^{- 4} m o l L^{- 1} s^{- 1}; 2.5 \times 1 0^{- 4} m o l L^{- 1} s^{- 1}$

$2.5 \times 1 0^{- 4} m o l L^{- 1} s^{- 1}; 7.5 \times 1 0^{- 4} m o l L^{- 1} s^{- 1}$

$2.5 \times 1 0^{- 4} m o l L^{- 1} s^{- 1}; 2.5 \times 1 0^{- 4} m o l L^{- 1} s^{- 1}$

$2.5 \times 1 0^{- 4} m o l L^{- 1} s^{- 1}; 5.0 \times 1 0^{- 4} m o l L^{- 1} s^{- 1}$

Q. The decomposition of NH3​ on platinum surface is zero order reaction. The rates of production of N2​ and H2​ are respectively (k=2.5×10−4molL−1s−1)

7.5×10−4molL−1s−1;2.5×10−4molL−1s−1

2.5×10−4molL−1s−1;7.5×10−4molL−1s−1

2.5×10−4molL−1s−1;2.5×10−4molL−1s−1

2.5×10−4molL−1s−1;5.0×10−4molL−1s−1

Q. The decomposition of $N H_{3}$ on platinum surface is zero order reaction. The rates of production of $N_{2}$ and $H_{2}$ are respectively $(k = 2.5 \times 1 0^{- 4} m o l L^{- 1} s^{- 1})$

$7.5 \times 1 0^{- 4} m o l L^{- 1} s^{- 1}; 2.5 \times 1 0^{- 4} m o l L^{- 1} s^{- 1}$

$2.5 \times 1 0^{- 4} m o l L^{- 1} s^{- 1}; 7.5 \times 1 0^{- 4} m o l L^{- 1} s^{- 1}$

$2.5 \times 1 0^{- 4} m o l L^{- 1} s^{- 1}; 2.5 \times 1 0^{- 4} m o l L^{- 1} s^{- 1}$

$2.5 \times 1 0^{- 4} m o l L^{- 1} s^{- 1}; 5.0 \times 1 0^{- 4} m o l L^{- 1} s^{- 1}$