Answer:
The enthalpy of the reaction is coming out to be -380.16 kJ.
Explanation:
Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as 
The equation used to calculate enthalpy change is of a reaction is:
![\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H_f%28product%29%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H_f%28reactant%29%5D)
For the given chemical reaction:
The equation for the enthalpy change of the above reaction is:
![\Delta H_{rxn}=[(2 mol\times \Delta H_f_{(N_2O)})+(2 mol\times\Delta H_f_{(H_2O)} )]-[(1 mol\times \Delta H_f_{(N_2H_4)})+(1 mol\times \Delta H_f_{(N_2O_4)})]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5B%282%20mol%5Ctimes%20%5CDelta%20H_f_%7B%28N_2O%29%7D%29%2B%282%20mol%5Ctimes%5CDelta%20H_f_%7B%28H_2O%29%7D%20%29%5D-%5B%281%20mol%5Ctimes%20%5CDelta%20H_f_%7B%28N_2H_4%29%7D%29%2B%281%20mol%5Ctimes%20%5CDelta%20H_f_%7B%28N_2O_4%29%7D%29%5D)
We are given:
Putting values in above equation, we get:
![\Delta H_{rxn}=[(2 mol\times 81.6 kJ/mol)+2 mol\times -241.8 kJ/mol)]-[(1 mol\times (50.6 kJ/mol))+(1 mol\times (9.16))]\\\\\Delta H_{rxn}=-380.16 kJ](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5B%282%20mol%5Ctimes%2081.6%20kJ%2Fmol%29%2B2%20mol%5Ctimes%20-241.8%20kJ%2Fmol%29%5D-%5B%281%20mol%5Ctimes%20%2850.6%20kJ%2Fmol%29%29%2B%281%20mol%5Ctimes%20%289.16%29%29%5D%5C%5C%5C%5C%5CDelta%20H_%7Brxn%7D%3D-380.16%20kJ)
Hence, the enthalpy of the reaction is coming out to be -380.16 kJ.
Answer:
What I came up with was D equaling out to be 42
Explanation:
Im assuming you just do the math
A. 3.5 x 104 = 364
B. 2.4 x 103 = 247.2
C. 1.4 x 10^2_2 = 280
D. 4.9 x 10-7 = 42
I hope I helped.
Answer:
<span>Formula for sodium hydroxide is represented by <u>NaOH</u>.
Explanation:
The reaction of Sodium oxide with water is as follow,
Na</span>₂O + H₂O → 2 NaOH
<span>
Na</span>₂O is considered as a strong basic oxide as it contains O²⁻ which has high tendency to bind with hydrogen atoms. This reaction is an exothermic reaction and is conducted in cold water to produce NaOH.<span>
</span>
Please help me answer your questions about how I feel you have a good time to carbon dinosaurs in the past and
Answer:
Option A
Explanation:
A) Yes. The reaction reaches equilibrium when the rate of reaction of the reverse reaction is equal to the rate of the forward reaction , then the only cause for the reverse reaction to be favoured is that the initial rate of the reverse was greater than the forward one.
B) No. The rate constant of the reverse reaction can be greater than the forward one but the rate also depends on concentrations, thus a reverse reaction with greater rate constant can result in the net reaction proceeding in the forward reaction, the reverse reaction or be at equilibrium depending on the concentrations or reactants and products
C) No. A lower activation energy means a higher rate constant , but a higher rate constant does not mean that the net reaction will proceed to the reactants ( see point B)
D) No. The energy changes determine conditions under thermodynamic equilibrium and therefore the net direction of the reaction will depend on the temperature and concentrations of reactants and products with respect to the equilibrium conditions.
