Answer:
Option B. 2096.1 K
Explanation:
Data obtained from the question include the following:
Enthalpy (H) = +1287 kJmol¯¹ = +1287000 Jmol¯¹
Entropy (S) = +614 JK¯¹mol¯¹
Temperature (T) =.?
Entropy is related to enthalphy and temperature by the following equation:
Change in entropy (ΔS) = change in enthalphy (ΔH) / Temperature (T)
ΔS = ΔH / T
With the above formula, we can obtain the temperature at which the reaction will be feasible as follow:
ΔS = ΔH / T
614 = 1287000/ T
Cross multiply
614 x T = 1287000
Divide both side by 614
T = 1287000/614
T = 2096.1 K
Therefore, the temperature at which the reaction will be feasible is 2096.1 K
Answer:
Explanation:
![\Delta H_{rxn}=\sum [n_{i}\times \Delta H_{f}^{0}(product)_{i}]-\sum [n_{j}\times \Delta H_{f}^{0}(reactant_{j})]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5Csum%20%5Bn_%7Bi%7D%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B0%7D%28product%29_%7Bi%7D%5D-%5Csum%20%5Bn_%7Bj%7D%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B0%7D%28reactant_%7Bj%7D%29%5D)
Where 
and 
are number of moles of product and reactant respectively (equal to their stoichiometric coefficient).
is standard heat of formation.
So, ![\Delta H_{rxn}=[2mol\times \Delta H_{f}^{0}(CO_{2})_{g}]+[3mol\times \Delta H_{f}^{0}(H_{2}O)_{g}]-[1mol\times \Delta H_{f}^{0}(C_{2}H_{5}OH)_{l}]-[3mol\times \Delta H_{f}^{0}(O_{2})_{g}]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5B2mol%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B0%7D%28CO_%7B2%7D%29_%7Bg%7D%5D%2B%5B3mol%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B0%7D%28H_%7B2%7DO%29_%7Bg%7D%5D-%5B1mol%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B0%7D%28C_%7B2%7DH_%7B5%7DOH%29_%7Bl%7D%5D-%5B3mol%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B0%7D%28O_%7B2%7D%29_%7Bg%7D%5D)
or, ![\Delta H_{rxn}=[2mol\times -393.509kJ/mol]+[3mol\times -241.818kJ/mol]-[1mol\times -277.69kJ/mol]-[3mol\times 0kJ/mol]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5B2mol%5Ctimes%20-393.509kJ%2Fmol%5D%2B%5B3mol%5Ctimes%20-241.818kJ%2Fmol%5D-%5B1mol%5Ctimes%20-277.69kJ%2Fmol%5D-%5B3mol%5Ctimes%200kJ%2Fmol%5D)
or,
Answer:
We need 27.56 moles hydrogen to produce 13.78 mol of ethane. (option 3)
Explanation:
Step 1: Data given
Moles ethane produced = 13.78 moles
Step 2: The balanced equation
C2H2 + 2H2 → C2H6
Step 3: Calculate moles of hydrogen
For 1 mol acetylene (C2H2) we need 2 moles hydrogen (H2) to produce 1 mol of ethane (C2H6)
For 13.78 moles ethane produced we need 2*13.78 = 27.56 moles hydrogen (H2)
We need 27.56 moles hydrogen to produce 13.78 mol of ethane. (option 3)
Producers(plants) produce organic compounds through photosynthesis. They are then eaten by secondary consumers in the next tropic level who obtain the energy the plants have. Then those animals are eaten by more secondary consumers in the next trophic level, obtaining more energy, and the cycle continues
The answer is _molecule_.
Hope this helps
