Answer:
Equilibrium constant Kc for the reaction will be 1.722
Explanation:
O2(g)+NO(g)→CO(g)+ NO2(g)
0.88 3.9 --- ---
0.88x 3.9-x x x
GIVEN:
0.88X-X= 0.11
⇒ X=0.77
CO2(g)+NO(g) → CO(g) + NO2(g)
0.88 3.9 --- ---
0.88-x 3.9-x x x
= 3.13 0.77 0.77
=0.11
Kc =
=
= 1.722
Answer:
The heat at constant pressure is -3,275.7413 kJ
Explanation:
The combustion equation is 2C₆H₆ (l) + 15O₂ (g) → 12CO₂ (g) + 6H₂O (l)
= (12 - 15)/2 = -3/2
We have;
Where R and T are constant, and ΔU is given we can write the relationship as follows;
Where;
H = The heat at constant pressure
U = The heat at constant volume = -3,272 kJ
= The change in the number of gas molecules per mole
R = The universal gas constant = 8.314 J/(mol·K)
T = The temperature = 300 K
Therefore, we get;
H = -3,272 kJ + (-3/2) mol ×8.314 J/(mol·K) ×300 K) × 1 kJ/(1000 J) = -3,275.7413 kJ
The heat at constant pressure, H = -3,275.7413 kJ.
Answer:
A. The pressure will increase 4 times. P₂ = 4 P₁
B. The pressure will decrease to half its value. P₂ = 0.5 P₁
C. The pressure will decrease to half its value. P₂ = 0.5 P₁
Explanation:
Initially, we have n₁ moles of a gas that occupy a volume V₁ at temperature T₁ and pressure P₁.
<em>What would happen to the gas pressure inside the cylinder if you do the following?</em>
<em />
<em>Part A: Decrease the volume to one-fourth the original volume while holding the temperature constant. Express your answer in terms of the variable P initial.</em>
V₂ = 0.25 V₁. According to Boyle's law,
P₁ . V₁ = P₂ . V₂
P₁ . V₁ = P₂ . 0.25 V₁
P₁ = P₂ . 0.25
P₂ = 4 P₁
<em>Part B: Reduce the Kelvin temperature to half its original value while holding the volume constant. Express your answer in terms of the variable P initial.</em>
T₂ = 0.5 T₁. According to Gay-Lussac's law,
<em>Part C: Reduce the amount of gas to half while keeping the volume and temperature constant. Express your answer in terms of the variable P initial.</em>
n₂ = 0.5 n₁.
P₁ in terms of the ideal gas equation is:
P₂ in terms of the ideal gas equation is: