Answer: The enthalpy change for this reaction is 392 Joules
Explanation:
The balanced chemical reaction is,
![H_2(g)+CO_2(g)\rightarrow H_2O(g)+CO(g)](https://tex.z-dn.net/?f=H_2%28g%29%2BCO_2%28g%29%5Crightarrow%20H_2O%28g%29%2BCO%28g%29)
The expression for enthalpy change is,
![\Delta H=\sum [n\times B.E_{reactant}]-\sum [n\times B.E_{product}]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Csum%20%5Bn%5Ctimes%20B.E_%7Breactant%7D%5D-%5Csum%20%5Bn%5Ctimes%20B.E_%7Bproduct%7D%5D)
![\Delta H=[(n_{H_2}\times B.E_{H_2})+(n_{CO_2}\times B.E_{CO_2})]-[(n_{H_2O}\times B.E_{H_2O})+(n_{CO}\times B.E_{CO})]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B%28n_%7BH_2%7D%5Ctimes%20B.E_%7BH_2%7D%29%2B%28n_%7BCO_2%7D%5Ctimes%20B.E_%7BCO_2%7D%29%5D-%5B%28n_%7BH_2O%7D%5Ctimes%20B.E_%7BH_2O%7D%29%2B%28n_%7BCO%7D%5Ctimes%20B.E_%7BCO%7D%29%5D)
where,
n = number of moles
Now put all the given values in this expression, we get
![\Delta H=[(1\times 1\times H-H)+(1\times 2\times C=O) ]-[(1\times 2\times O-H)+(2\times 1\times C-O)]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B%281%5Ctimes%201%5Ctimes%20H-H%29%2B%281%5Ctimes%202%5Ctimes%20C%3DO%29%20%5D-%5B%281%5Ctimes%202%5Ctimes%20O-H%29%2B%282%5Ctimes%201%5Ctimes%20C-O%29%5D)
![\Delta H=[(1\times 1\times 436)+(1\times 2\times 799) ]-[(1\times 2\times 463)+(2\times 1\times 358)]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B%281%5Ctimes%201%5Ctimes%20436%29%2B%281%5Ctimes%202%5Ctimes%20799%29%20%5D-%5B%281%5Ctimes%202%5Ctimes%20463%29%2B%282%5Ctimes%201%5Ctimes%20358%29%5D)
![\Delta H=392J](https://tex.z-dn.net/?f=%5CDelta%20H%3D392J)
Therefore, the enthalpy change for this reaction is 392 Joules
Answer:
0.21 M. (2 sig. fig.)
Explanation:
The molarity of a solution is the number of moles of the solute in each liter of the solution. The unit for molarity is M. One M equals to one mole per liter.
How many moles of NaOH in the original solution?
,
where
is the number of moles of the solute in the solution.
is the concentration of the solution.
for the initial solution.
is the volume of the solution. For the initial solution,
for the initial solution.
.
What's the concentration of the diluted solution?
.
is the number of solute in the solution. Diluting the solution does not influence the value of
.
for the diluted solution.- Volume of the diluted solution:
.
Concentration of the diluted solution:
.
The least significant number in the question comes with 2 sig. fig. Keep more sig. fig. than that in calculations but round the final result to 2 sig. fig. Hence the result: 0.021 M.
When we have this equation:
CO(g) + Cl2(g) ↔ COCl2(g)
intial 0.147 0.175 0
change -X -X +X
final (0.147-X) (0.175-X) X
so from the ICE table, we substitute in Kc formula :(when we have Kc = 255)
Kc = [COCl2]/[CO][Cl2]
255= X / (0.147-X)(0.175-X)
255 = X / (X^2 - 0.322 X + 0.025725)
X = 0.13
∴[CO] = 0.147 - X = 0.147 - 0.13
= 0.017 m
Answer:
Explanation:
a) In an exothermic reaction, the energy transferred to the surroundings from forming new bonds is ___more____ than the energy needed to break existing bonds.
b) In an endothermic reaction, the energy transferred to the surroundings from forming new bonds is ___less____ than the energy needed to break existing bonds.
c) The energy change of an exothermic reaction has a _____negative_______ sign.
d) The energy change of an endothermic reaction has a ____positive________ sign.
The energy changes occur during the bonds formation and bonds breaking.
There are two types of reaction endothermic and exothermic reaction.
Endothermic reactions:
The type of reactions in which energy is absorbed are called endothermic reactions.
In this type of reaction energy needed to break the bond are higher than the energy released during bond formation.
For example:
C + H₂O → CO + H₂
ΔH = +131 kj/mol
it can be written as,
C + H₂O + 131 kj/mol → CO + H₂
Exothermic reaction:
The type of reactions in which energy is released are called exothermic reactions.
In this type of reaction energy needed to break the bonds are less than the energy released during the bond formation.
For example:
Chemical equation:
C + O₂ → CO₂
ΔH = -393 Kj/mol
it can be written as,
C + O₂ → CO₂ + 393 Kj/mol
Answer:
I think b will be your answer
Explanation:
bc it makes a lot of sense