In the given situation, the reaction is-
NO + H2 ↔ Products
The rate of the reaction can be expressed (in terms of the decrease in the concentration of the reactants) as-
Rate = -dΔ[NO]/dt = -dΔ[H2]/dt
Now, if the concentration of NO is decreased there will be fewer molecules of the reactant NO which would decrease the its collision with H2. As a result the rate of the forward reaction would also decrease.
Ans) A decrease in the concentration of nitrogen monoxide decreases the collisions between NO and H2 molecules. the rate of the forward reaction then decreases.
Answer:Low temperatures
Explanation:
∆G= ∆H-T∆S
If ∆H is negative (exothermic reaction), then in order to maintain ∆G<0 which is the condition for spontaneity; T must decrease. This is because, decrease in T will keep the difference of ∆H and T∆S at a negative value in order to satisfy the above stated condition for spontaneity.
Answer:
Explanation:
From the correct question above:
The reaction can be represented as:
From the above reaction; the ICE table can be represented as:
I (mol/L) 0.086 0.28 0 0
C -4x -3x +2x +6x
E 0.086 - 4x 0.28 - 3x +2x +6x
At equilibrium;
The water vapor = 
![\text{equilibrium constant} ({k_c}) = \dfrac{ [N_2]^2 [H_2O]^6 }{ [[NH_3]^4] [O_2]^3 }](https://tex.z-dn.net/?f=%5Ctext%7Bequilibrium%20constant%7D%20%20%28%7Bk_c%7D%29%20%3D%20%20%5Cdfrac%7B%20%5BN_2%5D%5E2%20%5BH_2O%5D%5E6%20%7D%7B%20%5B%5BNH_3%5D%5E4%5D%20%5BO_2%5D%5E3%20%7D)
Replacing the value of x, we have:
Answer:
1.2×10² mmole of Na₂S₂O₃
Explanation:
From the question given above, the following data were obtained:
Volume = 0.6 L
Molarity = 0.2 mol/L
Mole of Na₂S₂O₃ =?
Molarity is simply defined as the mole of solute per unit litre of water. Mathematically, it is expressed as:
Molarity = mole /Volume
With the above formula, we can obtain the number of mole of Na₂S₂O₃ in the solution as illustrated below:
Volume = 0.6 L
Molarity = 0.2 mol/L
Mole of Na₂S₂O₃ =?
Molarity = mole /Volume
0.2 = Mole of Na₂S₂O₃ / 0.6
Cross multiply
Mole of Na₂S₂O₃ = 0.2 × 0.6
Mole of Na₂S₂O₃ = 0.12 mole
Finally, we shall convert 0.12 mole to millimole (mmol). This can be obtained as follow:
1 mole = 1000 mmol
Therefore,
0.12 mole = 0.12 mole × 1000 mmol / 1 mole
0.12 mole = 120 = 1.2×10² mmole
Thus, the chemist added 1.2×10² mmole of Na₂S₂O₃
