Answer:
The two ways to measure mass are subtraction and taring.
Explanation:
N2O4(g) <----------> 2NO2(g)
Before proceeding,
A chemical equilibrium can be defined as a condition in the course of a reversible chemical reaction in which no net change in the amounts of reactants and products occurs.
Statement 1.
This statement is false. Equilibrium is not about equal concentrations but rather zero change in concentration of the reactants and products.
Statement 2.
This statement is True in chemical equilibrium; the forward and reverse reactions occur at equal rates.
Statement 3.
This statement is False. The rate constant for the forward reaction is not equal to the rate constant of the reverse reaction.
Statement 4.
The concentration of NO2 divided by the concentration of N2O4 is NOT equal to a constant. To obtain a constant value irregardless of the concentrations, the concentration of NO2 must be squared. This comes from the stoichiometry of the reaction
Kc= [NO2]2 / [N2O4]
This statement is false.
Answer: The heat required is 6.88 kJ.
Explanation:
The conversions involved in this process are :
Now we have to calculate the enthalpy change.
![\Delta H=[m\times c_{p,s}\times (T_{final}-T_{initial})]+n\times \Delta H_{fusion}+[m\times c_{p,l}\times (T_{final}-T_{initial})]+n\times \Delta H_{vap}+[m\times c_{p,g}\times (T_{final}-T_{initial})]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Bm%5Ctimes%20c_%7Bp%2Cs%7D%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29%5D%2Bn%5Ctimes%20%5CDelta%20H_%7Bfusion%7D%2B%5Bm%5Ctimes%20c_%7Bp%2Cl%7D%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29%5D%2Bn%5Ctimes%20%5CDelta%20H_%7Bvap%7D%2B%5Bm%5Ctimes%20c_%7Bp%2Cg%7D%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29%5D)
where,
= enthalpy change = ?
m = mass of ethanol = 25.0 g
= specific heat of solid ethanol= 0.97 J/gK
= specific heat of liquid ethanol = 2.31 J/gK
n = number of moles of ethanol = 
= enthalpy change for fusion = 5.02 KJ/mole = 5020 J/mole
= change in temperature
The value of change in temperature always same in Kelvin and degree Celsius.
Now put all the given values in the above expression, we get
![\Delta H=[25.0 g\times 0.97J/gK\times (-114-(-135)K]+0.534mole\times 5020J/mole+[25.0g\times 2.31J/gK\times (-50-(-114))K]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B25.0%20g%5Ctimes%200.97J%2FgK%5Ctimes%20%28-114-%28-135%29K%5D%2B0.534mole%5Ctimes%205020J%2Fmole%2B%5B25.0g%5Ctimes%202.31J%2FgK%5Ctimes%20%28-50-%28-114%29%29K%5D)
(1 KJ = 1000 J)
Therefore, the heat required is 6.88 kJ
