finding enthalpy, molar enthalpy of vaporization, standard enthalpy of silicon tetrachloride, standard state of reaction

Question

Gallium 

 

e) After the spoon has melted, the puddle of liquid Gallium at the bottom of the coffee cup will continue to absorb energy until the Gallium and coffee reach thermal equilibrium.

 

If the initial temperature of the Gallium is the metal's melting point and the coffee's initial temperature is the one that you calculated in part d, what is the final temperature of the metal and coffee?

 

 

reaction

1. SiO2(s) +2C(s)---->Si(l) +2CO(g)

2. Si(s)+ 2Cl2(g)---->SiCl4(l)

3.SiCl4(g)+2Mg(s)--->2MgCl2(s)+Si(s)

Delta ReactionH= 738.28 kJ/mole-rxn

Delta FormationH=kJ/mole

a)what is the enthalpy of formation of liquid silicon?

b)what is the molar enthalpy of vaporization of liquid silicon tetrachloride in kj/mole at compound's boiling point temperature, 57.65 C

c)what is the standard state of enthalpy of formation of silicon tetrachloride, and what is the reaction that describes this process?

d)what is the standard state enthalpy of reaction, delta reaction H for Reaction #3

 

Answer 1

Continuing the upper part from a previous post (see previous post for notation details)

 

e)  Here, we can use a single expression to equate the heat transferred from the coffee into the liquid gallium, until the two bodies reach the same final temperature T_f'

 

Q_Ga + Q_cof = 0

 

This formula just means that all the heat transferred into the liquid gallium comes from the coffee.  Since the coffee has heat transfer our of itself to the liquid gallium, its heat transfer Q_cof is negative, which makes it so this equation can be true.

 

Assuming no phase changes to start, we can write:

 

Q_ga = m_gac_Ga-liquidΔT = m_ga c_Ga-liquid (T_f' - T_i-Ga)

Q_cof = m_cofc_cof(T_f'-T_i-cof)  (note:  this shows Q_cof to be negative, since we expect T_f' < T_i-cof)

 

We are given all the constants (in the previous post), and told that Tii-Ga = 29.76 ^oC (the melting temperature) and Ti-cof is the value from Part d), in the previous post (I obtained Ti-cof = 93.26 ^oC)

 

Using all these terms and values in the equation above, snd solving for Tf', I get Tf' = 92.81 ^oC.

 

This answers squares with a couple "health checks" (to make sure we did not go off the rails in calculation). 

 

1.  It is between the starting temperature of the coffee and the starting temperature of the gallium, which it should be.

2.  It is a LOT closer to the initial temperature of the coffee.  Since coffee has a decidedly higher specific heat, and there is a larger mass of it, we expect its temperature to change much more slowly with heat transfer than the liquid gallium does.  Therefore, the liquid gallium's temperature change should be much faster, and cover much more of the difference between the two initial temperatures.  The result bears this out.

 

If you want to talk in more detail about this or other problems of thermodynamics (or physics in general), I am available for online consultation.  Just let me know!

 

 

I cannot comment on the second question about the silicon, chlorine, and magnesium compounds, as I am not familiar enough with chemistry notation to know if the information is contained in there, or if it needs to be looked up.  I will leave that for another tutor.