35 x 75% + 37 x 25% = 35.5
Answer:
K = 8.1 x 10⁻³
Explanation:
We are told here that these gas phase reactions are both elementary processes, thus the reactions forward and reverse are both first order:
A→B Rate(forward) = k(forward) x [A]
and for
B→A Rate(reverse) = k(reverse) x [B]
At equilibrium we know the rates of the forward and reverse reaction are equal, so
k(forward) x [A] = k(reverse) x [B] for A(g)⇌B(g)
⇒ k(forward) / k(reverse) = [B] / [A] = K
4.7 x 10⁻³ s⁻1 / 5.8 x 10⁻¹ s⁻¹ = 8.1 x 10⁻³ = K
Notice how this answer is logical : the rate of the reverse reaction is greater than the forward reaction ( a factor of approximately 120 times) , and will be expecting a number for the equilibrium constant, K, smaller than one where the reactant concentration, [A], will prevail.
It is worth to mention that this is only valid for reactions which are single, elementary processes and not true for other equilibria.
Answer:
-196 kJ
Explanation:
By the Hess' Law, the enthalpy of a global reaction is the sum of the enthalpies of the steps reactions. If the reaction is multiplied by a constant, the value of the enthalpy must be multiplied by the same constant, and if the reaction is inverted, the signal of the enthalpy must be inverted too.
2S(s) + 3O₂(g) → 2SO₃(g) ΔH = -790 kJ
S(s) + O₂(g) → SO₂(g) ΔH = -297 kJ (inverted and multiplied by 2)
2S(s) + 3O₂(g) → 2SO₃(g) ΔH = -790 kJ
2SO₂(g) → 2S(s) + 2O₂(g) ΔH = +594 kJ
-------------------------------------------------------------
2S(s) + 3O₂(g) + 2SO₂(g) → 2SO₃(g) + 2S(s) + 2O₂(g)
Simplifing the compounds that are in both sides (bolded):
2SO₂(g) + O₂(g) → 2SO₃(g) ΔH = -790 + 594 = -196 kJ
Answer:
Explanation:
Significant figure implies number of digits that are to be considered. Some rules are required to be considered when writing a given expression to an expected significant figures.
So that:
1) 0.00004050 is 4 significant figures
2) 54.7000 is 6 significant figures
3) 1,000.09 is 6 significant figures
4) 0.039 is 2 significant figures
