Answer : The energy removed must be, -67.7 kJ
Solution :
The process involved in this problem are :
The expression used will be:
![\Delta H=[m\times c_{p,g}\times (T_{final}-T_{initial})]+m\times \Delta H_{vap}+[m\times c_{p,l}\times (T_{final}-T_{initial})]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Bm%5Ctimes%20c_%7Bp%2Cg%7D%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29%5D%2Bm%5Ctimes%20%5CDelta%20H_%7Bvap%7D%2B%5Bm%5Ctimes%20c_%7Bp%2Cl%7D%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29%5D)
where,
= heat released by the reaction = ?
m = mass of benzene = 125 g
= specific heat of gaseous benzene = 
= specific heat of liquid benzene = 
= enthalpy change for vaporization = 
Molar mass of benzene = 78.11 g/mole
Now put all the given values in the above expression, we get:
![\Delta H=[125g\times 1.06J/g.K\times (353.0-(425.0))K]+125g\times -434.0J/g+[125g\times 1.73J/g.K\times (335.0-353.0)K]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B125g%5Ctimes%201.06J%2Fg.K%5Ctimes%20%28353.0-%28425.0%29%29K%5D%2B125g%5Ctimes%20-434.0J%2Fg%2B%5B125g%5Ctimes%201.73J%2Fg.K%5Ctimes%20%28335.0-353.0%29K%5D)
Therefore, the energy removed must be, -67.7 kJ
Answer:
The limiting reactant is H₂
Explanation:
The reaction of hydrogen (H₂) and carbon monoxide (CO) to produce methanol (CH₃OH) is the following:
2H₂(g) + CO(g) → CH₃OH(g)
From the balanced chemical equation, we can see that 1 mol of CO reacts wIth 2 moles of H₂. So, the stoichiometric ratio is:
2 mol H₂/1 mol CO = 2.0
We have 500 mol of CO and 750 mol of H₂, so we calculate the ratio to establish a comparison:
750 mol H₂/500 mol CO = 1.5
Since 2.0 > 1.5, we have fewer moles of H₂ than are needed to completely react with 500 moles of CO. In fact, we need 1000 moles of H₂ and we have 750 moles. So, the limiting reactant is H₂.
Answer:
Rate of forward reaction will increase.
Explanation:
Effect of change in reaction condition on equilibrium is explained by Le Chatelier's principle. According to this principle,
If an equilibrium condition of a dynamic reversible reaction is disturbed by changing concentration, temperature, pressure, volume, etc, then reaction will move will in a direction which counteract the change.
In the given reaction,
A + B ⇌ C + D
If concentration of A is increase, then reaction will move in a direction which decreases the concentration of A to reestablish the equilibrium.
As concentration A decreases in forward direction, therefore, rate of forward reaction will increase.
Answer:
a
Explanation:just did the question and I guessed an answer cuz I didnt know it and I got it wrong so its a for a p e x
Answer:
The concentration of methyl isonitrile will become 15% of the initial value after 10.31 hrs.
Explanation:
As the data the rate constant is not given in this description, However from observing the complete question the rate constant is given as a rate constant of 5.11x10-5s-1 at 472k .
Now the ratio of two concentrations is given as
Here C/C_0 is the ratio of concentration which is given as 15% or 0.15.
k is the rate constant which is given as 
So time t is given as
So the concentration will become 15% of the initial value after 10.31 hrs.
