The standard molar free energies of formation of NO_2(g)andN_2o_4(g) at 25degree

Question

The standard molar free energies of formation of NO_2(g)andN_2o_4(g) at 25degree C are 51.840 and 98.280 kl/mol, respectively. What is the value of k_p(in atm) for the following equilibrium at 25degree c? 2NO_2 = N_2O_4? 0.113 8.84 1.17 *10^4 1.34 * 10^8 THe following questions refer to the following reaction at constant 25 degree c ans 1 atm. 2Fe(s) + O_2(g) + 2H_2O(1) - 2Fe(oh)_2(s) deltaH = -568.2 kj/mol Substance s degree (l/mol K) Fe(OH)_2(s) 80 Fe(s) 27 o_2(g) 205 H_2o(1) 70 Determine deltaSsurr for the reaction (in kj/mol K) 1.91 +3.14 -1.91 +0.36 Determine deltaSuniv for the above reaction (in kj/mol K) -2.22 +0.36 1.36 1.67 What must be true about deltaG for the above reaction? deltaG 0 deltag = TdeltaSuniv deltaG = 0 A stable diatomic molecule spontaneously forms from its atomes(all in gas states) in an exothermic manner: 2x(g) - x_2(g) What are the signs of deltaH degree, deltaS degree, and Deltag degree for this formation reaction? +, +, + -, -, + -,-,- +, +, -

Explanation / Answer

Solution :-

Q6)

                        2 NO2(g)   ------- > N2O4(g)

Delta Go        51.840 kJ                  98.280 kJ

Delta Go rxn = sum of delta G product - sum of delta g reactant

                     = [N2O4*1] – [NO2*2]

                     = [98.280*1] – [ 51.840 *2]

                     = -5.4 kJ

Now lets calculate the Kp

Delta Go = - RT ln Kp

(-5.4 kJ * 1000 J / 1kJ) = - 8.314 J per mol K * 298 K * ln Kp

(-5.4 kJ * 1000 J / 1kJ) / - 8.314 J per mol K * 298 K = ln Kp

2.18 = ln Kp

Anti ln 2.18 = kp

8.84 = Kp

So the answer is option B

Q7) 2Fe(s) +O2(g) + 2H2O(l) ------ > 2Fe(OH)2                   Delta H = -568.2 kJ/mol

Now lets calculate the delta S reaction

Delta S syst = Delta H / T

                   = -568.2 kJ / 298 K

                    = -1.91 kJ

Delta S rxn = sum of delta S product - sum of delta S reactant

                  = [ Fe(OH)2 * 2] - [(2*Fe) +(1*O2) +( 2*H2O)]

                 = [80*2] – [ (27*2)+(205*1)+(70*2)]

                 = -239 J / K

-239 J/K * 1 kJ / 1000 J = -0.239 kJ/K

Delta S surr = - Delta S sys

                    = -(-1.91 kJ)

                    = 1.91 kJ

So the answer is option A

Q8)Delta S univ = Delta S sys + delta S surr

                      = -0.239 kJ + 1.91 kJ

                      = +1.67 kJ

So the correct answer is option D

Q9) since the Delta H reaction is negative and therefore the delta G of the reaction will be negative

That is delta G <O

Q10) 2reactant gas atoms changes to 1 gas mole

So entropy decreasing

Enthalpy will be negative

And free energy will also be negative

So its

Option C

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