I believe the correct answer from the choices listed above is option A. <span>A forward reaction in which adding heat decreases product formation is exothermic, while a forward reaction in which adding heat increases product formation is endothermic. Exothermic would mean that heat is being released by the process while the opposite is called endothermic in which it absorbs heat.</span>
Answer:
See explanation below
Explanation:
The question is incomplete. However, here's the missing part of the question:
<em>"For the following reaction, Kp = 0.455 at 945 °C: </em>
<em>C(s) + 2H2(g) <--> CH4(g). </em>
<em>At equilibrium the partial pressure of H2 is 1.78 atm. What is the equilibrium partial pressure of CH4(g)?"</em>
With these question, and knowing the value of equilibrium of this reaction we can calculate the partial pressure of CH4.
The expression of Kp for this reaction is:
Kp = PpCH4 / (PpH2)²
We know the value of Kp and pressure of hydrogen, so, let's solve for CH4:
PpCH4 = Kp * PpH2²
*: You should note that we don't use Carbon here, because it's solid, and solids and liquids do not contribute in the expression of equilibrium, mainly because their concentration is constant and near to 1.
Now solving for PpCH4:
PpCH4 = 0.455 * (1.78)²
<u><em>PpCH4 = 1.44 atm</em></u>
Answer:
See Explanation
Explanation:
The equation of the reaction;
KHSO4(aq) + KOH(aq) -------> K2SO4(aq) + H2O(l)
Number of moles of KHSO4 = 49.6 g/136.169 g/mol = 0.36 moles
Since the reaction is in a mole ratio of 1:1, 0.36 moles of K2SO4 is produced.
Number of moles of KOH = 25.3 g/56.1056 g/mol = 0.45 moles
Since the reaction is 1:1, 0.45 moles of K2SO4 is produced
Hence K2SO4 is the limiting reactant.
Mass of K2SO4 formed = 0.36 moles of K2SO4 * 174.26 g/mol = 62.7 g
So;
1 mole of KHSO4 reacts with 1 mole of KOH
0.36 moles of KHSO4 reacts with 0.36 * 1/1 = 0.36 moles of KOH
Amount of excess KOH = 0.45 moles - 0.36 moles = 0.09 moles
Mass of excess KOH = 0.09 moles * 56.1056 g/mol = 5 g of excess KOH
