Question

In the reaction 4NH3+5O2 4NO+6H2O if the rate of disappearance of NH3 is 3.6*10 ki power -3 mol L*-1s*-1,what is the rate of formation of H2o

Answer 1

Answer:- The rate of formation of water is $5.4*10^-^3mol*L^-^1*s^-^1$ .

Solution:-  the given balanced equation is:

$4NH_3+5O_2\rightarrow 4NO+6H_2O$

From the balanced equation:-

$-\frac{1}{4}(\frac{\Delta NH_3}{\Delta t})=-\frac{1}{5}(\frac{\Delta O_2}{\Delta t})=\frac{1}{4}(\frac{\Delta NO}{\Delta t})=\frac{1}{6}(\frac{\Delta H_2O}{\Delta t})$

rate of disappearance of ammonia is given and we are asked to calculate the rate of formation of water. So, from the above relationship:

$-\frac{1}{4}(\frac{\Delta NH_3}{\Delta t})=\frac{1}{6}(\frac{\Delta H_2O}{\Delta t})$

It could be rearranged as:

$(\frac{\Delta H_2O}{\Delta t})=-\frac{6}{4}(\frac{\Delta NH_3}{\Delta t})$

rate of disappearance for ammonia is given, let's plug in it's value in the above equation:

$(\frac{\Delta H_2O}{\Delta t})=\frac{6}{4}(3.6*10^-^3mol*L^-^1*s^-^1)$

$(\frac{\Delta H_2O}{\Delta t})=5.4*10^-^3mol*L^-^1*s^-^1$

So, the rate of formation of water is $5.4*10^-^3mol*L^-^1*s^-^1$ .

Answer 2

Answer:

The rate of formation of water is $5.4\times 10^{-3}mol/L s$

Explanation:

$4NH_3+5O_2 \rightarrow 4NO+6H_2O$

Rate of disappearance $NH_3=-\frac{1}{4}\frac{d[NH_3]}{dt}=3.6\times 10^{-3} mol/L s$

Rate of formation of $H_2O$

$-\frac{1}{4}\frac{d[NH_3]}{dt}=\frac{1}{6}\frac{d[H_2O]}{dt}$

$3.6\times 10^{-3} mol/L s=\frac{1}{6}\frac{d[H_2O]}{dt}$

$\frac{d[H_2O]}{dt}=3.6\times 10^{-3} mol/L s\times \frac{6}{4} =5.4\times 10^{-3}mol/L s$

The rate of formation of water is $5.4\times 10^{-3}mol/L s$