Question

The chemical reaction for the formation of syngas is: CH4 + H2O -> CO + 3 H2 What is the rate for the formation of hydrogen, if the rate of the formation of carbon monoxide is 0.35 M/s ? g

Answer 1

Answer :  The rate for the formation of hydrogen is, 1.05 M/s

Explanation :

The general rate of reaction is,

$aA+bB\rightarrow cC+dD$

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

$\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}$

$\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}$

$\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}$

$\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}$

$Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}$

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

$CH_4+H_2O\rightarrow CO+3H_2$

The expression for rate of reaction :

$\text{Rate of disappearance of }CH_4=-\frac{d[CH_4]}{dt}$

$\text{Rate of disappearance of }H_2O=-\frac{d[H_2O]}{dt}$

$\text{Rate of formation of }CO=+\frac{d[CO]}{dt}$

$\text{Rate of formation of }H_2=+\frac{1}{3}\frac{d[H_2]}{dt}$

The rate of reaction expression is:

$\text{Rate of reaction}=-\frac{d[CH_4]}{dt}=-\frac{d[H_2O]}{dt}=+\frac{d[CO]}{dt}=+\frac{1}{3}\frac{d[H_2]}{dt}$

As we are given that:

$+\frac{d[CO]}{dt}=0.35M/s$

Now we to determine the rate for the formation of hydrogen.

$+\frac{1}{3}\frac{d[H_2]}{dt}=+\frac{d[CO]}{dt}$

$+\frac{1}{3}\frac{d[H_2]}{dt}=0.35M/s$

$\frac{d[H_2]}{dt}=3\times 0.35M/s$

$\frac{d[H_2]}{dt}=1.05M/s$

Thus, the rate for the formation of hydrogen is, 1.05 M/s