Answer:
C₇H₁₆ + 32CoF₃ —> C₇F₁₆ + 16HF + 32CoF₂
Explanation:
C₇H₁₆ + CoF₃ —> C₇F₁₆ + HF + CoF₂
The above equation can be balance as illustrated below:
C₇H₁₆ + CoF₃ —> C₇F₁₆ + HF + CoF₂
There are 16 atoms of H on the left side and 1 atom on the right side. It can be balance by writing 16 before HF as shown below:
C₇H₁₆ + CoF₃ —> C₇F₁₆ + 16HF + CoF₂
There are 3 atoms of F on the left side and a total of 34 atoms on the right side. It can be balance by writing 32 before CoF₃ and 32 before CoF₂ as shown below:
C₇H₁₆ + 32CoF₃ —> C₇F₁₆ + 16HF + 32CoF₂
Now, the equation is balanced.
Our reaction balanced equation at equilibrium N2(g) + 3 H2(g) ↔ 2 NH3(g)
and we have the Kp value at equilibrium = 4.51 X 10^-5
A) 98 atm NH3, 45 atm N2, 55 atm H2
when Kp = [P(NH3)]^2 / [P(N2)] * [P(H2)]^3
= 98^2 / (45 * 55^3) = 1.28 x 10^-3
by comparing the Kp by the Kp at equilibrium(the given value) So,
Kp > Kp equ So the mixture is not equilibrium,
it will shift leftward (to decrease its value) towards the reactants to achieve equilibrium.
B) 57 atm NH3, 143 atm N2, no H2
∴ Kp = [P(NH3)]^2 / [P(N2)]
= 57^2 / 143 = 22.7
∴Kp> Kp equ (the given value)
∴it will shift leftward (to decrease its value) towards reactants to achieve equilibrium.
c) 13 atm NH3, 27 atm N2, 82 atm H2
∴Kp = [P(NH3)]^2 / [P(N2)] * [P(H2)]^3
= 13^2 / (27* 82^3) = 1.14 X 10^-5
∴ Kp< Kp equ (the given value)
∴it will shift rightward (to increase its value) towards porducts to achieve equilibrium.
That would be either, nuclear power or wind power, depending on what the teacher is looking for...
this equation does support the law of conservation of mass.