The temperature of a reaction

In summary: I will fix it now.In summary, the equilibirum temperature is calculated using K=0.187, ΔS0=940J, and ΔH0=1414kJ. However, the equilibrium temperature is not calculated using standard conditions.
  • #1
Nikitin
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I'm supposed to find out the temperature of a reaction at equilibrium. I've calculated K=0.187, ΔS0=940J, and I know that ΔG0=1134kJ and ΔH0=1414kJ.

I'm supposed to use the R*T*ln(K)=ΔG0 formula, but my only problem is that ΔG0 is given for standard conditions,, ie where the temperature equals 298K.

But I am supposed to use it anyway in the assignment, ie I need to reform it into ΔG0=ΔH0 + T*ΔS0=T*R*ln(K), and calculate T.

However, ΔS and ΔH for the reaction are given at standard conditions. When the temperature changes they too change, so how is it possible to get a reliable answer?
 
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  • #2
help?
 
  • #3
Can you post the original question? You might be missing some kind of assumption, like the temperature remains constant for the reaction.
 
  • #4
Here's the reaction. Ca3(PO4)2 (s) + 3 SiO2 (s) + 5 C (s) -> 3 CaSiO3 (s) + 1/2 P4 (g) + 5 CO (g)

I'm supposed to find the equilibrium temperature, with the data from the OP (all of which are at standard conditions). I am also to assume that none of the substances change their aggregate states (ie none of them turn into gases) going from standard temperature to the equilibrium temperature.
 
  • #5
Can you find out K at standard conditions. Using this, relate K and T at standard conditions with K and T asked in the question by Van't Hoff equation.[tex]\mathrm{ln}\frac{K_2}{K_1}=\frac{-\Delta H^{\circ}}{R}\left ( \frac{1}{T_2}-\frac{1}{T_1} \right )[/tex]
 
  • #6
Do you really need K? Isn't it just a matter of finding T such that [itex]\Delta G = \Delta H - T \Delta S = 0[/itex]?
 
  • #7
AGNuke: I knew K at equil. temperature.. What I was struggling with was assuming that ΔS and ΔH are the same at 298K and 1500K for the reaction... Anyway I feel I understand allot more.

Borek said:
Do you really need K? Isn't it just a matter of finding T such that [itex]\Delta G = \Delta H - T \Delta S = 0[/itex]?
The assignment wanted that the equilibrum temperature be calculated for K=0.187.. ugh I was probably unclear about this.. sorry I'm sleep deprived as heck.

EDIT: When I wrote that ΔG0=R*T*ln(K) in the OP, I meant ΔG0=-R*T*ln(K). I forgot the minus-sign infront of R*T*ln(K).
 
Last edited:

What is the temperature of a reaction?

The temperature of a reaction refers to the amount of heat energy within the reaction system. It is a measure of the average kinetic energy of the particles involved in the reaction.

Why is the temperature of a reaction important?

The temperature of a reaction plays a crucial role in determining the rate and outcome of a chemical reaction. It affects the speed at which reactants collide, the activation energy required for the reaction to occur, and the equilibrium constant of the reaction.

How is the temperature of a reaction measured?

The temperature of a reaction is typically measured using a thermometer or a temperature probe. The instrument is placed in the reaction mixture and the temperature is recorded at regular intervals throughout the reaction.

How does the temperature affect the rate of a reaction?

An increase in temperature generally leads to an increase in the rate of a reaction. This is because a higher temperature provides more energy for particles to collide, increasing the chances of successful collisions and therefore, the rate of the reaction.

What happens if the temperature of a reaction is too high?

If the temperature of a reaction becomes too high, the reaction may proceed too quickly, leading to uncontrollable or even dangerous conditions. This could result in the formation of unwanted byproducts or even explosions. It is important to carefully control and monitor the temperature of a reaction to ensure safe and efficient results.

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