2NaCl + Br2
This is because Na (Sodium) still needs to retain the coefficient or amount of two. Since Chlorine already has 2 on the reactants side, having a two in front would make sense so both elements can have a two.
Bromine would have the subscript of two as in the reactants side of the equation, it is also under the coefficient of two. Thus, it would need to carry it to the products side in order to keep the equation balanced.
Hope this helps!
Answer:
![\Delta H_{rxn}=-2043.999kJ](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D-2043.999kJ)
Explanation:
![\Delta H_{rxn}^{0}=\sum [n_{i}\times \Delta H_{f}^{0}(product)_{i}]-\sum [n_{j}\times \Delta H_{f}^{0}(reactant_{j})]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%5E%7B0%7D%3D%5Csum%20%5Bn_%7Bi%7D%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B0%7D%28product%29_%7Bi%7D%5D-%5Csum%20%5Bn_%7Bj%7D%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B0%7D%28reactant_%7Bj%7D%29%5D)
Where
and
are number of moles of product and reactant respectively (equal to their stoichiometric coefficient).
is standard heat of formation and
is standard enthalpy change for reaction at ![25^{0}\textrm{C}](https://tex.z-dn.net/?f=25%5E%7B0%7D%5Ctextrm%7BC%7D)
So, ![\Delta H_{rxn}=[3mol\times \Delta H_{f}^{0}(CO_{2})_{g}]+[4mol\times \Delta H_{f}^{0}(H_{2}O)_{g}]-[1mol\times \Delta H_{f}^{0}(C_{3}H_{8})_{g}]-[5mol\times \Delta H_{f}^{0}(O_{2})_{g}]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5B3mol%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B0%7D%28CO_%7B2%7D%29_%7Bg%7D%5D%2B%5B4mol%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B0%7D%28H_%7B2%7DO%29_%7Bg%7D%5D-%5B1mol%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B0%7D%28C_%7B3%7DH_%7B8%7D%29_%7Bg%7D%5D-%5B5mol%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B0%7D%28O_%7B2%7D%29_%7Bg%7D%5D)
or, ![\Delta H_{rxn}=[3mol\times -393.509kJ/mol]+[4mol\times -241.818kJ/mol]-[1mol\times -103.8kJ/mol]-[5mol\times 0kJ/mol]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5B3mol%5Ctimes%20-393.509kJ%2Fmol%5D%2B%5B4mol%5Ctimes%20-241.818kJ%2Fmol%5D-%5B1mol%5Ctimes%20-103.8kJ%2Fmol%5D-%5B5mol%5Ctimes%200kJ%2Fmol%5D)
or, ![\Delta H_{rxn}=-2043.999kJ](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D-2043.999kJ)
Answer:
[H2] = [I2] = 0.64M; [HI] = 4.72M
Explanation:
Based on the reaction:
H2 + I2 ⇄ 2HI
The K is defined as:
k = 54.3 = [HI]² / [H2] [I2]
<em>Where [] is molar concentration of each reactant at equilibrium</em>
As the initial concentration of HI is 6mol/dm^3 = 6M the equilibrium concentration of each reactant is:
[H2] = X
[I2] = X
[HI] = 6 - 2X
<em>Where X is reaction coordinate</em>
<em />
Replacing:
54.3 = [6-2X]² / [X] [X]
54.3X² = 4X² - 24X + 36
0 = -50.3X² - 24X + 36
Solving for X:
X = -1.12. False solution, produce negative concentrations
X = 0.64M. Right solution
Replacing:
[H2] = 0.64M
[I2] = 0.64M
[HI] = 6 - 2*0.64M = 4.72M
Equilibrium concentrations are:
<h3>[H2] = [I2] = 0.64M; [HI] = 4.72M</h3>