Answer : The correct option is, (C) spontaneous only at low temperatures.
Explanation :
According to Gibb's equation:
= Gibbs free energy
= enthalpy change
= entropy change
T = temperature in Kelvin
As we know that:
= +ve, reaction is non spontaneous
= -ve, reaction is spontaneous
= 0, reaction is in equilibrium
The given chemical reaction is:
As we are given that, the given reaction is exothermic that means the enthalpy change is negative.
In this reaction, the randomness of reactant molecules are more and as we move towards the formation of product the randomness become less that means the degree of disorderedness decreases. So, the entropy will also decreases that means the change in entropy is negative.
Now we have to determine the spontaneity of this reaction when ΔH is negative and ΔS is negative.
As,
(at high temperature) (non-spontaneous)
(at low temperature) (spontaneous)
Thus, the reaction is spontaneous only at low temperatures.
Answer:
4e-5 is the length in meters
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Answer:
![K_a=\frac{[H_3O^+][HCO_3^-]}{[H_2CO_3]}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7B%5BH_3O%5E%2B%5D%5BHCO_3%5E-%5D%7D%7B%5BH_2CO_3%5D%7D)
Explanation:
Several rules should be followed to write any equilibrium expression properly. In the context of this problem, we're dealing with an aqueous equilibrium:
- an equilibrium constant is, first of all, a fraction;
- in the numerator of the fraction, we have a product of the concentrations of our products (right-hand side of the equation);
- in the denominator of the fraction, we have a product of the concentrations of our reactants (left-hand side o the equation);
- each concentration should be raised to the power of the coefficient in the balanced chemical equation;
- only aqueous species and gases are included in the equilibrium constant, solids and liquids are omitted.
Following the guidelines, we will omit liquid water and we will include all the other species in the constant. Each coefficient in the balanced equation is '1', so no powers required. Multiply the concentrations of the two products and divide by the concentration of carbonic acid:
126.751 g/mol
Formula: FeCl2
Ca=40
N=14
O=16
Ca=40
N=14x2=28
O=6x16=96
molar mass=40+28+96
molar mass=164g/mol
