For the reaction 2NH3 → N2+3H2, if -(d[NH3]/dt)=k1[NH3], (d [N2]/dt)=k2[NH3], (d[H2]/dt)=K3[NH3] then the relation between k1, k2 and k3 is

Question

For the reaction 2NH3 → N2+3H2, if -(d[NH3]/dt)=k1[NH3], (d [N2]/dt)=k2[NH3], (d[H2]/dt)=K3[NH3] then the relation between k1, k2 and k3 is (A) k1 = k2 = k3 (B) k1 = 3k2 = 2k3 (C) 1.5 k1 = 3k2 = k3 (D) 2k1 = k2 = 3k3

Tardigrade Admin · Accepted Answer

2NH3 arrow N2+3H2 Rate = -(1/2)(d[NH3]/dt)=(d[N2]/dt)=(1/3)(d[H2]/dt) (1/2)k1[NH3]=k2[NH3]=(1/3)k3[NH3] 1.5k1=3k2=k3

Q. For the reaction $2 N H_{3} \to N_{2} + 3 H_{2}$ ,
if $- \frac{d [ N H _{3} ]}{d t} = k_{1} [N H_{3}], \frac{d [ N _{2} ]}{d t} = k_{2} [N H_{3}], \frac{d [ H _{2} ]}{d t} = K_{3} [N H_{3}]$
then the relation between $k_{1}, k_{2}$ and $k_{3}$ is

$k_{1} = k_{2} = k_{3}$

$k_{1} = 3 k_{2} = 2 k_{3}$

$1.5 k_{1} = 3 k_{2} = k_{3}$

$2 k_{1} = k_{2} = 3 k_{3}$

$2 N H_{3} \to N_{2} + 3 H_{2}$
Rate $= - \frac{1}{2} \frac{d [ N H _{3} ]}{d t} = \frac{d [ N _{2} ]}{d t} = \frac{1}{3} \frac{d [ H _{2} ]}{d t}$
$\frac{1}{2} k_{1} [N H_{3}] = k_{2} [N H_{3}] = \frac{1}{3} k_{3} [N H_{3}]$
$1.5 k_{1} = 3 k_{2} = k_{3}$

Q. For the reaction 2NH3​→N2​+3H2​, if −dtd[NH3​]​=k1​[NH3​],dtd[N2​]​=k2​[NH3​],dtd[H2​]​=K3​[NH3​] then the relation between k1​,k2​ and k3​ is

k1​=k2​=k3​

k1​=3k2​=2k3​

1.5k1​=3k2​=k3​

2k1​=k2​=3k3​