Question

A) Find the gas speed of ammonia at 25.0 degrees Celsius. ______________

Answer 1

a. 661.23 m/s

b.  the rate of effusion of Ammonia = 4.5 faster than Silicon tetra bromide

Further explanation

Given

T = 25 + 273 = 298 K

Required

a. the gas speed

b. The rate of effusion comparison

Solution

a.

Average velocities of gases can be expressed as root-mean-square averages. (V rms)  

$\large {\boxed {\bold {v_ {rms} = \sqrt {\dfrac {3RT} {Mm}}}}$

R = gas constant, T = temperature, Mm = molar mass of the gas particles  

From the question  

R = 8,314 J / mol K  

T = temperature  

Mm = molar mass, kg / mol  

Molar mass of Ammonia = 17 g/mol = 0.017 kg/mol

$\tt v=\sqrt{\dfrac{3\times 8.314\times 298}{0.017} }=661.23~m/s$

b. the effusion rates of two gases = the square root of the inverse of their molar masses:  

$\rm \dfrac{r_1}{r_2}=\sqrt{\dfrac{M_2}{M_1} }$

M₁ = molar mass Ammonia NH₃= 17

M₂ =  molar mass Silicon tetra bromide SiBr₄= 348

$\rm \dfrac{r_1}{r_2}=\sqrt{\dfrac{348}{17} }=4.5$

the rate of effusion of Ammonia = 4.5 faster than Silicon tetra bromide