A _________________ is developed through the scientific method, and it can be modified or improved upon. It may be represented b
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This is an incomplete question, here is a complete question.
Consider the following equilibrium at 100°C.
Concentration at equilibrium:
If a system has a reaction quotient of 2.13 × 10⁻¹⁵ at 100°c, what will happen to the concentrations of COBr₂, Co and Br₂ as the reaction proceeds to equilibrium?
Answer : The concentrations of Co and Br₂ decreases and the concentrations of COBr₂ increases.
Explanation :
Reaction quotient (Q) : It is defined as the measurement of the relative amounts of products and reactants present during a reaction at a particular time.
The given balanced chemical reaction is,
The expression for reaction quotient will be :
In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.
Now put all the given values in this expression, we get
The given equilibrium constant value is,
Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.
There are 3 conditions:
When that means product > reactant. So, the reaction is reactant favored.
When that means reactant > product. So, the reaction is product favored.
When that means product = reactant. So, the reaction is in equilibrium.
From the above we conclude that, the that means product < reactant. So, the reaction is product favored that means reaction must shift to the product (right) to be in equilibrium.
Hence, the concentrations of Co and Br₂ decreases and the concentrations of COBr₂ increases.
Answer: The mass of the gas is 18.3 g/mol.
Explanation:
To calculate the rate of diffusion of gas, we use Graham's Law.
This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows:
Squaring both sides and solving for
Hence, the molar mas of unknown gas is 18.3 g/mol.
Answer:
0.2g
Explanation:
Given parameters:
Number of moles of H₂ = 0.4mol
Number of moles of O₂ = 0.15mol
Unknown:
Mass of reactant that would remain = ?
Solution:
To solve this problem, we need to know the limiting reactant which is the one in short supply in the given reaction.
The expression of the reaction is :
2H₂ + O₂ → 2H₂O
2 mole of H₂ will combine with 1 mole of O₂
But given; 0.4 mole of H₂ we will require = 0.2mole of O₂
The given number of oxygen gas is 0.15mole and it is the limiting reactant.
Hydrogen gas is in excess;
1 mole of oxygen gas will combine with 2 mole of hydrogen gas
0.15 mole of oxygen gas will require 0.15 x 2 = 0.3mole of hydrogen gas
Now, the excess mole of hydrogen gas = 0.4 mole - 0.3 mole = 0.1mole
Mass of hydrogen gas = number of mole x molar mass
Molar mass of hydrogen gas = 2(1) = 2g/mol
Mass of hydrogen gas = 0.1 x 2 = 0.2g