Develop cost efficient methods to use solar energy
increase the use of wind power
reduce the use of coal and oil
Answer:
1. Adding hydrogen gas, b. shift to the right
2. Adding a catalyst, c. No effect
3. Decreasing the pressure, a. shift to the left
Explanation:
Hydrogen gas can be rewritten as H2. Whenever you add something to an equilibrium expression, it will shift to whichever side does not have this. So, since the reactant side has 3 moles of H2, adding more H2 to the reaction will shift to the products side, since there is no H2 there.
Adding a catalyst has no effect on equilibrium reactions.
When decreasing the pressure, equilibrium will shift to the side with the greater number of moles of gas. In this case, there are 4 moles of gas on the left, and 2 on the right, so it would shift to the left.
<u>Answer:</u> The standard Gibbs free energy of the given reaction is 6.84 kJ
<u>Explanation:</u>
For the given chemical equation:
The expression of 
for above equation follows:
We are given:
Putting values in above expression, we get:
To calculate the equilibrium constant (at 25°C) for given value of Gibbs free energy, we use the relation:
where,
= standard Gibbs free energy = ?
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 = 0.0632
Putting values in above equation, we get:
Hence, the standard Gibbs free energy of the given reaction is 6.84 kJ
Answer:
0.3mol C8H18
Explanation:
For this we must first look at the reaction taking place:
C8H18+O2 --> H2O + CO2
Balancing the equation we get:
2(C8H18)+25(O2) --> 18(H2O) + 16(CO2)
Form there we now need to know how many moles of Octane are needed to produce 2.4moles of H2O. The conversion is as follows:
2.4molH2O ((2mol of C8H18)/(18mol of H2O)) = 0.3mol C8H18
