Respuesta :
Solution: The given balanced equation is:
[tex]4NH_3+5O_2\rightarrow 4NO+6H_2O[/tex]
If we have a hypothetical equation:
[tex]A+2B\rightarrow 3C+5D[/tex]
Then the rate could be written as:
[tex]rate=-\frac{\Delta [A]}{\Delta t}=-\frac{1}{2}\frac{\Delta [B]}{\Delta t}=\frac{1}{3}\frac{\Delta [C]}{\Delta t}=\frac{1}{5}\frac{\Delta [D]}{\Delta t}[/tex]
From above expression one thing could easily be noticed that the coefficients of all are inverted. Also, there is negative sign in front of reactants and positive sign in front of the products. Negative sign stands for rate of consumption where as positive sign stands for rate of formation.
Like the above example, we can write the rate for the given equation and it would be looking as:
[tex]rate=-\frac{1}{4}\frac{\Delta [NH_3]}{\Delta t}=-\frac{1}{5}\frac{\Delta [O_2]}{\Delta t}=\frac{1}{4}\frac{\Delta [NO]}{\Delta t}=\frac{1}{6}\frac{\Delta [H_2O]}{\Delta t}[/tex]
Now we can easily answer all the parts of the question.
(a) From above expression, the rate of consumption of [tex]O_2[/tex] related to rate of consumption of [tex]NH_3[/tex] as:
[tex]-\frac{1}{5}\frac{\Delta [O_2]}{\Delta t}=-\frac{1}{4}\frac{\Delta [NH_3]}{\Delta t}[/tex]
multiply both sides by -5
[tex]\frac{\Delta [O_2]}{\Delta t}=\frac{5}{4}\frac{\Delta [NH_3]}{\Delta t}[/tex]
So, rate of consumption of oxygen is [tex]\frac{5}{4}[/tex] the rate of consumption of ammonia.
(b) The relationship between rate of formation of NO to the rate of consumption of ammonia will be written as:
[tex]\frac{1}{4}\frac{\Delta [NO]}{\Delta t}=-\frac{1}{4}\frac{\Delta [NH_3]}{\Delta t}[/tex]
Multiply both sides by 4
[tex]\frac{\Delta [NO]}{\Delta t}=-\frac{\Delta [NH_3]}{\Delta t}[/tex]
So, rate of formation of NO equals to the rate of consumption of ammonia.
Now, the rate of formation of [tex]H_2O[/tex] to the rate of consumption of ammonia would be:
[tex]\frac{1}{6}\frac{\Delta [H_2O]}{\Delta t}=-\frac{1}{4}\frac{\Delta [NH_3]}{\Delta t}[/tex]
Multiply both sides by 6
[tex]\frac{\Delta [H_2O]}{\Delta t}=-\frac{6}{4}\frac{\Delta [NH_3]}{\Delta t}[/tex]
So, the rate of formation of [tex]H_2O[/tex] is [tex]\frac{6}{4}[/tex] times that is 1.5 times to the rate of consumption of ammonia.
