Calculating Enthalpy and Entropy for H2+I2--> 2HI Reaction at 718 K

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The equilibrium constant Kp for the reaction H2 + I2 → 2HI at 718 K is given as 50.2, and the standard enthalpy change (ΔH) is 25 kJ mol^-1. To calculate the entropy change (ΔS) at 718 K, the relationship between Gibbs free energy (ΔG), enthalpy, and entropy must be applied using the equation ΔG = ΔH - TΔS. The standard Gibbs free energy can be derived from Kp, allowing for the determination of ΔS at the specified temperature. The calculation can be formatted using LaTeX for clarity and precision.
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Kp for the reaction H2+I2--> 2HI is 50.2 is 718 K. The ▲H (enthalpy) on standard conditions is 25 kJ mol^-1 .Use this information to find the enthropy of the reaction on 718 K...so I can findf free Gibbs energy from Kp,(in standard conditions) and therefore the standard enthropy..but how about enthropy in 718 K?
 
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Could you show what you did? it will look nicer if you write it with latex :)
 
Thread 'Confusion regarding a chemical kinetics problem'

Thread 'Confusion regarding a chemical kinetics problem'

TL;DR Summary: cannot find out error in solution proposed. [![question with rate laws][1]][1] Now the rate law for the reaction (i.e reaction rate) can be written as: $$ R= k[N_2O_5] $$ my main question is, WHAT is this reaction equal to? what I mean here is, whether $$k[N_2O_5]= -d[N_2O_5]/dt$$ or is it $$k[N_2O_5]= -1/2 \frac{d}{dt} [N_2O_5] $$ ? The latter seems to be more apt, as the reaction rate must be -1/2 (disappearance rate of N2O5), which adheres to the stoichiometry of the...
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