Respuesta :
Answer :
The rate of change of [tex]NH_3[/tex] partial pressure is 126 torr/h.
The total pressure in the vessel is zero.
Explanation :
The general rate of reaction is,
[tex]aA+bB\rightarrow cC+dD[/tex]
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 :
[tex]\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}[/tex]
[tex]\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}[/tex]
[tex]\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}[/tex]
[tex]\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}[/tex]
[tex]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}[/tex]
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,
[tex]N_2H_4(g)+H_2(g)\rightarrow 2NH_3(g)[/tex]
The expression for rate of reaction :
[tex]\text{Rate of disappearance of }N_2H_4=-\frac{d[N_2H_4]}{dt}[/tex]
[tex]\text{Rate of disappearance of }H_2=-\frac{d[H_2]}{dt}[/tex]
[tex]\text{Rate of formation of }NH_3=+\frac{1}{2}\frac{d[NH_3]}{dt}[/tex]
As we know that, the partial pressure is directly proportional to the concentration. So,
[tex]\text{Rate of reaction}=-\frac{dP_{N_2H_4}}{dt}=-\frac{dP_{H_2}}{dt}=+\frac{1}{2}\frac{dP_{NH_3}}{dt}[/tex]
Given:
[tex]-\frac{dP_{N_2H_4}}{dt}=63torr/h[/tex]
As,
[tex]-\frac{dP_{N_2H_4}}{dt}=-\frac{dP_{H_2}}{dt}=63torr/h[/tex]
and,
[tex]-\frac{dP_{N_2H_4}}{dt}=+\frac{1}{2}\frac{dP_{NH_3}}{dt}[/tex]
[tex]63torr/h=+\frac{1}{2}\frac{dP_{NH_3}}{dt}[/tex]
[tex]\frac{dP_{NH_3}}{dt}=63torr/h\times 2=126torr/h[/tex]
Thus, the rate of change of [tex]NH_3[/tex] partial pressure is 126 torr/h.
Now we have to calculate the total pressure in the vessel.
[tex]\frac{dP_{Total}}{dt}=\frac{dP_{N_2H_4}}{dt}+\frac{dP_{H_2}}{dt}+\frac{1}{2}\frac{dP_{NH_3}}{dt}[/tex]
[tex]\frac{dP_{Total}}{dt}=(-63torr/h)+(-63torr/h)+(126torr/h)[/tex]
[tex]\frac{dP_{Total}}{dt}=0[/tex]
Thus, the total pressure in the vessel is zero.
The rate of change of the partial pressure of ammonia is 126 torr/h.
What is rate of reaction?
The rate of reaction is defined as the rate at which the reactants are decreased or the rate at which the products are increased. Given the equation; N2H4 (g) + H2 (g) -----> 2 NH3 (g), we know that;
-d[N2H4]/dt = -d[H2]/dt = 1/2d[NH3]/dt
It the follows that; -d[N2H4]/dt = -d[H2]/dt = 63 torr/h
Therefore;
d[NH3]/dt = 2 * 63 torr/h = 126 torr/h
Rtse of change of the total pressure = (-d[N2H4]/dt) + ( -d[H2]/dt) + 1/2d[NH3]/dt =- 63 torr/h - 63 torr/h + 126 torr/h = 0
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