Question

somebody solve this problem...(4.3)​

Answer 1

SoluTion :-

The order of the reaction is zero

Rate = 2.5 × 10⁻⁴ × 1 = 2.5 10⁻⁴ mol L⁻¹ S⁻¹

$\mathrm {\frac{d}{dt}\ N_{2}=\frac{1}{3} \frac{d}{dt} \ H_{2} }$

The rate of formation of N₂ = 2.5 × 10⁻⁴ mol L⁻¹ S⁻¹

$\sf {2.5 \times 10^{-4}=\frac{1}{3} \frac{d}{dt} \ H_{2}}\\\\\\\\\sf {\frac{d}{dt} \ H_{2}=7.5 \times 10^{-4}}$

Rate of formation of H₂ = 7.5 × 10⁻⁴ mol L⁻¹ S⁻¹