Question

For the reaction C6H14(g) ------> C6H6(g) + 4H2(g), the rate of formation of hydrogen gas, H2 was found to be 2.5 x 10-2 atm/s, measured in terms of pressure change. Determine the rate of consumption of hexane, C6H14(g) at the same time.

Answer 1

Answer : The rate of consumption of hexane is, $6.25\times 10^{-3}atm/s$

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,

$C_6H_{14}(g)\rightarrow C_6H_6(g)+4H_2(g)$

The expression for rate of reaction :

$\text{Rate of disappearance of }C_6H_{14}=-\frac{d[C_6H_{14}]}{dt}$

$\text{Rate of formation of }C_6H_6=+\frac{d[C_6H_6]}{dt}$

$\text{Rate of formation of }H_2=+\frac{1}{4}\frac{d[H_2]}{dt}$

As we know that,  the partial pressure is directly proportional to the concentration. So,

$\text{Rate of reaction}=-\frac{dP_{C_6H_{14}}}{dt}=+\frac{dP_{C_6H_6}}{dt}=+\frac{1}{4}\frac{dP_{H_2}}{dt}$

Given:

$+\frac{dP_{H_2}}{dt}=2.5\times 10^{-2}atm/s$

As,  

$-\frac{dP_{C_6H_{14}}}{dt}=+\frac{1}{4}\frac{dP_{H_2}}{dt}=2.5\times 10^{-2}atm/s$

and,

$\frac{dP_{C_6H_{14}}}{dt}=\frac{1}{4}\times 2.5\times 10^{-2}atm/s$

$\frac{dP_{C_6H_{14}}}{dt}=6.25\times 10^{-3}atm/s$

Thus, the rate of consumption of hexane is, $6.25\times 10^{-3}atm/s$