Answer:
More products (Fe2O3 and SO2) will be produced
Explanation:
In the reaction:
4 FeS2(s) + 11 O2(g) ⇌ 2 Fe2O3(s) + 8 SO2(g)
FeS2 is a reactant. When a reactant is added to a reaction which has already reached the equilibrium, more product is produced, so that the equilibrium is reestablished again
Answer:
1. ![R=k[A]^1[B]^2](https://tex.z-dn.net/?f=R%3Dk%5BA%5D%5E1%5BB%5D%5E2)
2. ![R=k[B]^1](https://tex.z-dn.net/?f=R%3Dk%5BB%5D%5E1)
3. ![R=k[A]^0[B]^0=k](https://tex.z-dn.net/?f=R%3Dk%5BA%5D%5E0%5BB%5D%5E0%3Dk)
4. ![R=k[A]^1[B]^{-1}](https://tex.z-dn.net/?f=R%3Dk%5BA%5D%5E1%5BB%5D%5E%7B-1%7D)
Explanation:
Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.
(1) is second order in B and overall third order.
2A + B → C
Order of the reaction = sum of stoichiometric coefficient
= x + 2 = 3
x = 1
Rate of the reaction =R
![R=k[A]^1[B]^2](https://tex.z-dn.net/?f=R%3Dk%5BA%5D%5E1%5BB%5D%5E2)
(2) is zero order in A and first order in B.
2A + B → C
Rate of the reaction =R
![R=k[A]^0[B]^1=k[B]^1](https://tex.z-dn.net/?f=R%3Dk%5BA%5D%5E0%5BB%5D%5E1%3Dk%5BB%5D%5E1)
Order of the reaction = sum of stoichiometric coefficient
= 0 + 1 = 1
(3) is zero order in both A and B .
2A + B → C
Order of the reaction = sum of stoichiometric coefficient
= 0 + 0 = 0
Rate of the reaction =R
![R=k[A]^0[B]^0=k](https://tex.z-dn.net/?f=R%3Dk%5BA%5D%5E0%5BB%5D%5E0%3Dk)
(4) is first order in A and overall zero order.
2A + B → C
Order of the reaction = sum of stoichiometric coefficient
= 1 + x = 0
x = -1
Rate of the reaction = R
![R=k[A]^1[B]^{-1}](https://tex.z-dn.net/?f=R%3Dk%5BA%5D%5E1%5BB%5D%5E%7B-1%7D)
Volume = ?
Molarity (M) = 1.00 x 10⁻² M
moles (n) = 3.00 x 10⁻¹ mol
V = n / M
V = 3.00 x 10⁻¹ / 1.00 x 10⁻²
V = 30 L
The correct answer should be "longer than".