The rate of decrease in n2h4 partial pressure in a closed reaction vessel form the reaction: n2h4 (g) + h2 (g) ightarrow 2 nh3 (g) is63 torr/h. what are the rates of change of nh3 partial pressure and total pressure in the vessel.

Answer :The rate of change of partial pressure is 126 torr/h.The total pressure in the vessel is ze...

Chemistry, 07.12.2019 00:31 kaylarenee05080

The rate of decrease in n2h4 partial pressure in a closed reaction vessel form the reaction: n2h4 (g) + h2 (g) \rightarrow 2 nh3 (g) is63 torr/h. what are the rates of change of nh3 partial pressure and total pressure in the vessel.

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Answer from: grandpianograce

Answer :

The rate of change of NH_3 partial pressure is 126 torr/h.

The total pressure in the vessel is zero.

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,

$N_2H_4(g)+H_2(g)\rightarrow 2NH_3(g)$

The expression for rate of reaction :

$\text{Rate of disappearance of }N_2H_4=-\frac{d[N_2H_4]}{dt}$

$\text{Rate of disappearance of }H_2=-\frac{d[H_2]}{dt}$

$\text{Rate of formation of }NH_3=+\frac{1}{2}\frac{d[NH_3]}{dt}$

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

$\text{Rate of reaction}=-\frac{dP_{N_2H_4}}{dt}=-\frac{dP_{H_2}}{dt}=+\frac{1}{2}\frac{dP_{NH_3}}{dt}$

Given:

$-\frac{dP_{N_2H_4}}{dt}=63torr/h$

As,

$-\frac{dP_{N_2H_4}}{dt}=-\frac{dP_{H_2}}{dt}=63torr/h$

and,

$-\frac{dP_{N_2H_4}}{dt}=+\frac{1}{2}\frac{dP_{NH_3}}{dt}$

$63torr/h=+\frac{1}{2}\frac{dP_{NH_3}}{dt}$

$\frac{dP_{NH_3}}{dt}=63torr/h\times 2=126torr/h$

Thus, the rate of change of NH_3 partial pressure is 126 torr/h.

Now we have to calculate the total pressure in the vessel.

$\frac{dP_{Total}}{dt}=\frac{dP_{N_2H_4}}{dt}+\frac{dP_{H_2}}{dt}+\frac{1}{2}\frac{dP_{NH_3}}{dt}$

$\frac{dP_{Total}}{dt}=(-63torr/h)+(-63torr/h)+(126torr/h)$

$\frac{dP_{Total}}{dt}=0$

Thus, the total pressure in the vessel is zero.

Answer from: Quest

definition

entropy is a measure of the randomness or the extent of disorder of a chemical process.

enthalpy is a measure of the heat change of a reaction occurring at a constant pressure.

measurement units:

entropy is measured in joule per kelvin (jk⁻¹) whereas enthalpy is measured in joule per mole (jmol⁻¹)

requirements:

entropy has no requirements or limits, and its change is measured by the division between the heat change of the chemical process and the temperature.

in contrast, enthalpy can only be related to the heat change of the reaction under standard conditions.

Answer from: Quest

The correct answer is b.hoped i you

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The rate of decrease in n2h4 partial pressure in a closed reaction vessel form the reaction: n2h4 (...