In a catalytic conversion of N2 to NH3 by Haber’s process, the rate of reaction was expressed as change in the concentration of ammonia per time is 40 × 10-3 mol L-1 s-1. If there are no side reaction, the rate of the reaction as expressed in terms of hydrogen is (in mol L-1 s-1)

Question

In a catalytic conversion of N2 to NH3 by Haber’s process, the rate of reaction was expressed as change in the concentration of ammonia per time is 40 × 10-3 mol L-1 s-1. If there are no side reaction, the rate of the reaction as expressed in terms of hydrogen is (in mol L-1 s-1) (A) 60 × 10-3 (B) 20 × 10-3 (C) 1.200 (D) 10.3 × 10-3

Tardigrade Admin · Accepted Answer

(-d(N2)/dt)=-(1/3)(d(H2)/dt)=(1/2)(d(NH3)/dt) =(d(H2)/dt)=(3/2)(d(NH3)/dt)=(3/2)×40×10-3 =60×10-3

$60 \times 1 0^{- 3}$

$20 \times 1 0^{- 3}$

$1.200$

$10.3 \times 1 0^{- 3}$

$\frac{- d ( N _{2} )}{d t} = - \frac{1}{3} \frac{d ( H _{2} )}{d t} = \frac{1}{2} \frac{d ( N H _{3} )}{d t}$
$= \frac{d ( H _{2} )}{d t} = \frac{3}{2} \frac{d ( N H _{3} )}{d t} = \frac{3}{2} \times 40 \times 1 0^{- 3}$
$= 60 \times 1 0^{- 3}$

Q. In a catalytic conversion of N2​ to NH3​ by Haber’s process, the rate of reaction was expressed as change in the concentration of ammonia per time is 40×10−3molL−1s−1. If there are no side reaction, the rate of the reaction as expressed in terms of hydrogen is (inmolL−1s−1)

60×10−3

20×10−3

1.200

10.3×10−3

dt−d(N2​)​=−31​dtd(H2​)​=21​dtd(NH3​)​ =dtd(H2​)​=23​dtd(NH3​)​=23​×40×10−3 =60×10−3

$60 \times 1 0^{- 3}$

$20 \times 1 0^{- 3}$

$1.200$

$10.3 \times 1 0^{- 3}$

$\frac{- d ( N _{2} )}{d t} = - \frac{1}{3} \frac{d ( H _{2} )}{d t} = \frac{1}{2} \frac{d ( N H _{3} )}{d t}$
$= \frac{d ( H _{2} )}{d t} = \frac{3}{2} \frac{d ( N H _{3} )}{d t} = \frac{3}{2} \times 40 \times 1 0^{- 3}$
$= 60 \times 1 0^{- 3}$