The standard enthalpy and entropy are taken from the Kirchhoff equation in integral form. The heat capacity is assumed constant.

Dear Audrey, you can spend the eternity waiting for this reaction to occur with liquid water at ambient pressure (i.e., under 373K). You'd better take H2O(g) instead and calculate Gibbs energy change for this reaction as function of temperature. It would be better to regard Cp as function of temperature, not a constant. For deltaG, you have to calculate two integrals: deltaH = integral(deltaCp)dT and deltaS = integral(deltaCp/T)dT. After that, you can caluclate deltaG = deltaH - T*deltaS. Cp for reagents and products you can pick from NIST Webbook (either tabulated or as Shomate equation) or from some physical chemistry handbook.

Thank you Andrei Novkov and Yuri Migorod. for your response.

Andrei Novikov i know this reaction can't proceed at atmpospheric prseeure and 373K, however by heating liquid water at about 700K the reaction can occur. My problem is how do i calculate the enthalpy and entropy of this reaction at 700K ( at 700K, liquid water vaporizes). So which enthalpy/entropy of formation (H2O(l) or H2O(g))at 298K do i need to use to evaluate the enthalpy.entropy of this reaction at 700K.

700 K is above the critical temperature of water, consequently the liquid state can no more exist.

You must use only thermodynamic data for gasous watere available for a very wide range of temperatures and pressures.

Thank you Stefan Perisanu.

This is what i did to calculate the entropy and enthalpy of this reaction at 700K.

From these datas under standard conditions

substance deltafHm deltaSm deltafGm Cp,m

(J.mol-1) ( J.mol-1 K-1) (J.mol-1) ( J.mol-1 K-1)

C(s) 0 5.7 0 8.5

H2O(l) -285800 69.9 -237100 75.3

H2O(g) -241800 188.8 -228600 33.6

CO(g) -110500 197.7 -137200 29.1

H2(g) 0 130.7 0 28.8

Given that Cp,m is independent of temperature and the vaporization enthalpy for H2O(l), deltavapH=40.7KJ.mol-1 at standard pressure and 373K.

C(s) + H2O(l) --> CO(g) +H2(g)

At 298K, 1atm;

deltaGreaction = sum of deltaGformation of products - sum of delta Gformation of reactants.

deltaGreaction = -137200+237100

= 99.9KJ.mol-1 ( since G> 0) this reaction is not spontaneous.

deltaHreaction = sum of deltaHformation of products - sum of delta Hformation of reactants.

deltaHreaction = -110500+285800

= 175.3KJ.mol-1

deltaSreaction = sum of deltaS of products - sum of deltaS of reactants.

deltaSreaction = 130.7+197.7 - (5.7+69.9)

= 252.8J.mol-1 K-1

Since, deltaG= deltaH - TdeltaS, and for a reaction to be spontaneous,

deltaG < 0;

so, to find the minimum temperature required to drive the reaction

deltaH - TdeltaS < 0

175.3KJ.mol-1 - T*252.8J.mol-1 K-1 < 0

and thus T> 694K for the reaction to proceed.

To evaluate the enthalpy and entropy of this reaction at 700K, this is what i did;

At 700K, 1atm

delta Hreaction = delta Hreaction(298K) + Cpreaction(298K )*(373K-298K)+ delta Hvap of H2O at 373K + Cpreaction(700K)*( 700K-373K).

Cp reaction(298K) = 29.1+28.8-(8.5+75.3) = -25.9J.mol-1 K-1

Cp reaction(700K) = 29.1+28.8-(8.5+33.6) = 15.8 J.mol-1 K-1 [ is this correct if Cp is independent of temperature?]

deltaHreaction = 175300 - 25.9(75) + 40700+ 15.8(327)

= 219.2KJmol-1

deltaSreaction = deltaSreaction(298K) + Cpreaction(298K )*(ln373K/298K) + delta Hvap of H2O/373K + Cpreaction(700K)*(ln700K/373K).

deltaSreaction = 252.8 -25.9ln(373/298) + 40700/373 + 15.8ln(700/373)

= 366.3J.mol-1 K-1

[ I will like to have your comments and suggestions on my approach ]