Answer:
a. Kc = 0,116
b. 1,11 atm
c. 25,5 g of NOBr
Explanation:
a. For the equilibrium:
2NOBr(g)⇌2NO(g)+Br₂(g)
kc is defined as:
<em>(1)</em>
The molar concentration of each compound are:
NOBr: 3,29 g×
= 0,0299 moles/ 5,00L = <em>5,986x10⁻³M</em>
[NO]: 3,04 g×
= 0,101 moles/ 5,00L = 0,0203M
[Br₂]: 8,10 g×
= 0,0507 moles/ 5,00L = <em>0,0101M</em>
Replacing in (1)
<em>kc = 0,116</em>
b. Total pressure could be obtained from the total molarity, thus:
P = MRT
Total molarity (M) is: 5,986x10⁻³M + 0,0203M + 0,0101M = 0,0364M
Thus, total pressure is:
P = 0,0364 mol/L × 0,082atmL/molK×373,15K -<em>100°C=373,15K-</em> = <em>1,11 atm</em>
c. Using the equilibrium it is possible to obtain initial moles of NOBr with moles of each compound in equilibrium and then, initial mass, thus:
0,101 moles NO×
= 0,101 mol NOBr
0,0507 moles Br₂×
= 0,101 mol NOBr
Thus, total initial moles of NOBr are:
0,101 mol + 0,101 mol + 0,0299 mol = 0,232 mol NOBr. In mass:
0,232 mol×
= <em>25,5 g of NOBr</em>
I hope it helps!