The given sentence is part of a longer question.
I found this question with the same sentence. So, I will help you using this question:
For the reaction N2O4<span>(g) ⇄ 2NO</span>2(g), a reaction mixture at a certain temperature initially contains both N2O4 and NO2 in their standard states (meaning they are gases with a pressure of 1 atm<span>). If </span>Kp = 0.15, which statement is true of the reaction mixture before
any reaction occurs?
(a) Q = K<span>; The reaction </span>is at equilibrium.
(b) Q < K<span>;
The reaction </span>will proceed to
the right.
(c) Q > K<span>; The reaction </span>will proceed to the left.
The answer is the option (c) Q > K<span>; The reaction will proceed to the </span>left,
since Qp<span> = </span>1<span>, and 1 > 0.15.</span>
Explanation:
Kp is the equilibrium constant in term of the partial pressures of the gases.
Q is the reaction quotient. It is a measure of the progress of a chemical reaction.
The reaction quotient has the same form of the equilibrium constant but using the concentrations or partial pressures at any moment.
At equilibrium both Kp and Q are equal. Q = Kp
If Q < Kp then the reaction will go to the right (forward reaction) trying to reach the equilibrium,
If Q > Kp then the reaction will go to the left (reverse reaction) trying to reach the equilibrium.
Here, the state is that both pressures are 1 atm, so Q = (1)^2 / 1 = 1.
Since, Q = 1 and Kp = 0.15, Q > Kp and the reaction will proceed to the left.
The answer is B!! hope this helps(: an example is a zebra or a lion
Answer:
According to the Brønsted definition, an acid is a substance capable of donating a proton, and a base is a substance capable of accepting a proton. ... The species giving up the proton is HCl, an acid. The species accepting the proton is water, the base. The species Cl- is the conjugate base of HCl.
Answer:
3. V = 0.2673 L
4. V = 2.4314 L
5. V = 0.262 L
6. V = 2.224 L
Explanation:
3. assuming ideal gas:
∴ R = 0.082 atm.L/K.mol
∴ V1 = 225 L
∴ T1 = 175 K
∴ P1 = 150 KPa = 1.48038 atm
⇒ n = RT/PV
⇒ n = ((0.082 atm.L/K.mol)(175 K))/((1.48038 atm)(225 L))
⇒ n = 0.043 mol
∴ T2 = 112 K
∴ P2 = P1 = 150 KPa = 1.48038 atm
⇒ V2 = RT2n/P2
⇒ V2 = ((0.082 atm.L/K.mol)(112 K)(0.043 mol))/(1.48038 atm)
⇒ V2 = 0.2673 L
4. gas is heated at a constant pressure
∴ T1 = 180 K
∴ P = 1 atm
∴ V1 = 44.8 L
⇒ n = RT/PV
⇒ n = ((0.082 atm.L/K.mol)(180 K))/((1 atm)(44.8 L))
⇒ n = 0.3295 mol
∴ T2 = 90 K
⇒ V2 = RT2n/P
⇒ V2 = ((0.082 atm.L/K.mol)(90 K)(0.3295 mol))/(1 atm)
⇒ V2 = 2.4314 L
5. V1 = 200 L
∴ P1 = 50 KPa = 0.4935 atm
∴ T1 = 271 K
⇒ n = RT/PV
⇒ n = ((0.082 atm.L/K.mol)(271 K))/((0.4935 atm)(200 L))
⇒ n = 0.2251 mol
∴ P2 = 100 Kpa = 0.9869 atm
∴ T2 = 14 K
⇒ V2 = RT2n/P2
⇒ V2 = ((0.082 atm.L/K.mol)(14 K)(0.2251 mol))/(0.9869 atm)
⇒ V2 = 0.262 L
6.a) ∴ V1 = 24.6 L
∴ P1 = 10 atm
∴ T1 = 25°C = 298 K
⇒ n = RT/PV
⇒ n = ((0.082 atm.L/K.mol)(298 K))/((10 atm)(24.6 L))
⇒ n = 0.0993 mol
∴ T2 = 273 K
∴ P2 = 101.3 KPa = 0.9997 atm
⇒ V2 = RT2n/P2
⇒ V2 = ((0.082 atm.L/K.mol)(273 K)(0.0993 mol))/(0.9997 atm)
⇒ V2 = 2.224 L
