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:
[tex]kc = \frac{[Br_{2}][NO]^2{[NOBr]^2}[/tex] (1)
The molar concentration of each compound are:
NOBr: 3,29 g×[tex]\frac{1mol}{109,92 g}[/tex] = 0,0299 moles/ 5,00L = 5,986x10⁻³M
[NO]: 3,04 g×[tex]\frac{1mol}{30 g}[/tex] = 0,101 moles/ 5,00L = 0,0203M
[Br₂]: 8,10 g×[tex]\frac{1mol}{159,8 g}[/tex] = 0,0507 moles/ 5,00L = 0,0101M
Replacing in (1)
kc = 0,116
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 -100°C=373,15K- = 1,11 atm
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×[tex]\frac{2molNOBr}{2molNO}[/tex] = 0,101 mol NOBr
0,0507 moles Br₂×[tex]\frac{2molNOBr}{1molBr_{2}}[/tex] = 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×[tex]\frac{109,92g}{1mol}[/tex] = 25,5 g of NOBr
I hope it helps!
Answer:
a) Kc = 0.116
b) The total pressure is 1.143 atm
c) The mass of the original sample of NOBr is 14.43 g
Explanation:
Please look at the solution in the attached Word document
