Answer:
The work done is 2,907 J
Explanation:
Work is one of the forms of energy transmission between bodies. To perform a job, you must exert a force on a body and it moves. In other words, when a force is applied to a body and it moves, work is done.
The work is equal to the product of the force by the distance and by the cosine of the angle that exists between the direction of the force and the direction that the moving point or object travels:
W = F * d * cos α
In the International System of Units the Force is expressed in newtons and the distance in meters, so the work will have units of newtons. Meter = joules (J).
In this case:
Replacing:
W= 255 N* 11.4 m* cos 0
Solving:
W= 255 N* 11.4 m* 1
W= 2,907 J
<u><em>The work done is 2,907 J</em></u>
I think it would be A. base pairs but that is my best answer Have a great day!
<u>Answer:</u> The equilibrium constant for this reaction is ![5.85\times 10^{5}](https://tex.z-dn.net/?f=5.85%5Ctimes%2010%5E%7B5%7D)
<u>Explanation:</u>
The equation used to calculate standard Gibbs free change is of a reaction is:
![\Delta G^o_{rxn}=\sum [n\times \Delta G^o_{(product)}]-\sum [n\times \Delta G^o_{(reactant)}]](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo_%7Brxn%7D%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20G%5Eo_%7B%28product%29%7D%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20G%5Eo_%7B%28reactant%29%7D%5D)
For the given chemical reaction:
![3H_2(g)+N_2(g)\rightarrow 2NH_3(g)](https://tex.z-dn.net/?f=3H_2%28g%29%2BN_2%28g%29%5Crightarrow%202NH_3%28g%29)
The equation for the standard Gibbs free change of the above reaction is:
![\Delta G^o_{rxn}=[(2\times \Delta G^o_{(NH_3(g))})]-[(1\times \Delta G^o_{(N_2)})+(3\times \Delta G^o_{(H_2)})]](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo_%7Brxn%7D%3D%5B%282%5Ctimes%20%5CDelta%20G%5Eo_%7B%28NH_3%28g%29%29%7D%29%5D-%5B%281%5Ctimes%20%5CDelta%20G%5Eo_%7B%28N_2%29%7D%29%2B%283%5Ctimes%20%5CDelta%20G%5Eo_%7B%28H_2%29%7D%29%5D)
We are given:
![\Delta G^o_{(NH_3(g))}=-16.45kJ/mol\\\Delta G^o_{(N_2)}=0kJ/mol\\\Delta G^o_{(H_2)}=0kJ/mol](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo_%7B%28NH_3%28g%29%29%7D%3D-16.45kJ%2Fmol%5C%5C%5CDelta%20G%5Eo_%7B%28N_2%29%7D%3D0kJ%2Fmol%5C%5C%5CDelta%20G%5Eo_%7B%28H_2%29%7D%3D0kJ%2Fmol)
Putting values in above equation, we get:
![\Delta G^o_{rxn}=[(2\times (-16.45))]-[(1\times (0))+(3\times (0))]\\\\\Delta G^o_{rxn}=-32.9kJ/mol](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo_%7Brxn%7D%3D%5B%282%5Ctimes%20%28-16.45%29%29%5D-%5B%281%5Ctimes%20%280%29%29%2B%283%5Ctimes%20%280%29%29%5D%5C%5C%5C%5C%5CDelta%20G%5Eo_%7Brxn%7D%3D-32.9kJ%2Fmol)
To calculate the equilibrium constant (at 25°C) for given value of Gibbs free energy, we use the relation:
![\Delta G^o=-RT\ln K_{eq}](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo%3D-RT%5Cln%20K_%7Beq%7D)
where,
= standard Gibbs free energy = -32.9 kJ/mol = -35900 J/mol (Conversion factor: 1 kJ = 1000 J )
R = Gas constant = 8.314 J/K mol
T = temperature = ![25^oC=[273+25]K=298K](https://tex.z-dn.net/?f=25%5EoC%3D%5B273%2B25%5DK%3D298K)
= equilibrium constant at 25°C = ?
Putting values in above equation, we get:
![-32900J/mol=-(8.314J/Kmol)\times 298K\times \ln K_{eq}\\\\K_{eq}=e^{13.279}=5.85\times 10^{5}](https://tex.z-dn.net/?f=-32900J%2Fmol%3D-%288.314J%2FKmol%29%5Ctimes%20298K%5Ctimes%20%5Cln%20K_%7Beq%7D%5C%5C%5C%5CK_%7Beq%7D%3De%5E%7B13.279%7D%3D5.85%5Ctimes%2010%5E%7B5%7D)
Hence, the equilibrium constant for this reaction is ![5.85\times 10^{5}](https://tex.z-dn.net/?f=5.85%5Ctimes%2010%5E%7B5%7D)