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klemol [59]
3 years ago
10

Consider the following reaction: CO(g)+2H2(g)⇌CH3OH(g) Kp=2.26×104 at 25 ∘C. Calculate ΔGrxn for the reaction at 25 ∘C under eac

h of the following conditions. PCH3OH= 1.4 atm ; PCO=PH2= 1.2×10−2 atm
Chemistry
1 answer:
valentinak56 [21]3 years ago
3 0

Answer : The value of \Delta G_{rxn} is, 8.867kJ/mole

Explanation :

The formula used for \Delta G_{rxn} is:

\Delta G_{rxn}=\Delta G^o+RT\ln Q   ............(1)

where,

\Delta G_{rxn} = Gibbs free energy for the reaction

\Delta G_^o =  standard Gibbs free energy

R = gas constant = 8.314 J/mole.K

T = temperature = 25^oC=273+25=298K

Q = reaction quotient

First we have to calculate the \Delta G_^o.

Formula used :

\Delta G^o=-RT\times \ln K_p

Now put all the given values in this formula, we get:

\Delta G^o=-(8.314J/mole.K)\times (298K)\times \ln (2.26\times 10^{4})

\Delta G^o=-24839.406J/mole=-24.83\times 10^3J/mole=-24.83kJ/mole

Now we have to calculate the value of 'Q'.

The given balanced chemical reaction is,

CO(g)+2H_2(g)\rightarrow CH_3OH(g)

The expression for reaction quotient will be :

Q=\frac{(p_{CH_3OH})}{(p_{CO})\times (p_{H_2})^2}

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

Now put all the given values in this expression, we get

Q=\frac{(1.4)}{(1.2\times 10^{-2})\times (1.2\times 10^{-2})^2}

Q=8.1\times 10^{5}

Now we have to calculate the value of \Delta G_{rxn} by using relation (1).

\Delta G_{rxn}=\Delta G^o+RT\ln Q

Now put all the given values in this formula, we get:

\Delta G_{rxn}=-24.83kJ/mole+(8.314\times 10^{-3}kJ/mole.K)\times (298K)\ln (8.1\times 10^{5})

\Delta G_{rxn}=8.867\times 10^3J/mole=8.867kJ/mole

Therefore, the value of \Delta G_{rxn} is, 8.867kJ/mole

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Answer:

             %age Yield  =  51.45 %

Solution:

Step 1: Convert Kg into g

68.5 Kg CO  =  68500 g CO

8.60 Kg H₂  =  8600 g

Step 2: Find out Limiting reactant;

The Balance Chemical Equation is as follow;

                                 CO  +  2 H₂    →    CH₃OH

According to Equation,

                   28 g (1 mol) CO reacts with  =  4 g (2 mol) of H₂

So,

                    68500 g CO will react with  =  X g of H₂

Solving for X,

                    X  =  (68500 g × 4 g) ÷ 28 g

                    X  =  9785 g of H₂

It shows 9785 g H₂ is required to react with 68500 g of CO but we are provided with 8600 g of H₂ which is less than required. Therefore, H₂ is provided in less amount hence, it is a Limiting reagent and will control the yield of products.

Step 3: Calculate Theoretical Yield

According to equation,

            4 g (2 mol) H₂ reacts to produce  =  32 g (1 mol) Methanol

So,

                          8600 g H₂ will produce  =  X g of CH₃OH

Solving for X,

                    X  =  (8600 g × 32 g) ÷ 4 g

                     X =  68800 g of CH₃OH

Step 4: Calculate %age Yield

                     %age Yield  =  Actual Yield ÷ Theoretical Yield × 100

Putting Values,

                     %age Yield  =  3.54 × 10⁴ g ÷ 68800 g × 100

                     %age Yield  =  51.45 %


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Explanation:

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Debemos multiplicar 160.000 por 1 y luego dividir entre 10 ;

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