Question

is 4.44 × 10-2 M/s, what is the rate of disappearance of NO? 2 NO(g) + Cl2(g) → 2 NOCl(g) Given the following balanced equation, determine the rate of reaction with respect to [Cl2]. If the rate of disappearance of Cl2 is 4.44 × 10-2 M/s, what is the rate of disappearance of NO? 2 NO(g) + Cl2(g) → 2 NOCl(g) 2.22 × 10-2 M/s 1.11 × 10-1 M/s 4.44 × 10-2 M/s 8.88 × 10-2 M/s 5.25 × 10-2 M/s

Answer 1

Answer: The rate of disappearance of NO is $8.88\times 10^{-2}M/s$

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.

$2NO(g)+Cl_2(g)\rightarrow 2NOCl(g)$

Rate=$-\frac{1d[NO]}{2dt}=-\frac{d[Cl_2]}{dt}=+\frac{1d[NOCl]}{2dt}$

Given: $\frac{-d[Cl_2]}{dt}]=4.44\times 10^{-2}M/s$

Thus the rate of disappearance of NO = $-\frac{1d[NO]}{dt}=2\times {\frac{-d[Cl_2]}{dt}=2\times 4.44\times 10^{-2}M/s=8.88\times 10^{-2}M/s$

The rate of disappearance of NO  is $8.88\times 10^{-2}M/s$