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4 years ago

14

In a first-order reaction, how does the rate change if the concentration of the reactant decreases to one-third its original val

ue?

2 answers:

garik1379 [7]4 years ago

3 0

The rate decreases by a factor of one third

podryga [215]4 years ago

3 0

Changes at the rate of 3 or the third rate

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Phosphorus trichloride gas and chlorine gas react to form phosphorus pentachloride gas: PCl3(g)+Cl2(g)⇌PCl5(g). A 7.5-L gas vess

Tpy6a [65]

Answer:

The equilibrium constant in terms of concentration that is, $K_c=3.6243\times 10^{3}$ .

Explanation:

$PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)$

The relation of $K_c\& K_p$ is given by:

$K_p=K_c(RT)^{\Delta n_g}$

$K_p$ = Equilibrium constant in terms of partial pressure.=98.1

$K_c$ = Equilibrium constant in terms of concentration =?

T = temperature at which the equilibrium reaction is taking place.

R = universal gas constant

$\Delta n_g$ = Difference between gaseous moles on product side and reactant side= $n_{g,p}-n_{g.r}=1-2=-1$

$98.1=K_c(RT)^{-1}$

$98.1 =\frac{K_c}{RT}$

$K_c=98.1\times 0.0821 L atm/mol K\times 450 K=3,624.30=3.6243\times 10^{3}$

The equilibrium constant in terms of concentration that is, $K_c=3.6243\times 10^{3}$ .

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3 years ago

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What type of element is most abundant: metals, nonmetals, or metalloids?

AnnyKZ [126]

Metals are the most abundant

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4 years ago

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An electrochemical cell has the following standard cell notation.

icang [17]

The cell notation is:

$Mg(s)|Mg^{+2}(aq)||Ag^{+}(aq)|Ag(s)$

here in cell notation the left side represent the anodic half cell where right side represents the cathodic half cell

in anodic half cell : oxidation takes place [loss of electrons]

in cathodic half cell: reduction takes place [gain of electrons]

1) this is a galvanic cell

2) the standard potential of cell will be obtained by subtracting the standard reduction potential of anode from cathode

$E^{0}_{Mg}=-2.38V$

$E^{0}_{Ag}=+0.80V$

Therefore

$E^{0}_{cell}=0.80-(-2.38)=+3.18V$

3) as the value of emf is positive the reaction will be spontaneous as the free energy change of reaction will be negative

Δ $G^{0}=-nFE^{0}$

As reaction is spontaneous and there will be conversion of chemical energy to electrical energy it is a galvanic cell.

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3 years ago

Which statement is true about metals and metalloids? A. All metals and metalloids have luster. B. Metals are good conductors and

Aleks [24]

I believe the statement that is true of metals and metalloids is B. Metals are good conductors and metalloids are bad conductors.

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3 years ago

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At a certain temperature, 0.760 mol SO3 is placed in a 1.50 L container. 2SO3(g) = 2SO2(g) + O2(g)At equilibrium, 0.130 mol O2 i

olganol [36]

Answer:

$K_c=0.0867$

Explanation:

Moles of SO₃ = 0.760 mol

Volume = 1.50 L

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$Molarity=\frac{0.760}{1.50\ L}$

[SO₃] = 0.5067 M

Considering the ICE table for the equilibrium as:

$\begin{matrix} & 2SO_3_{(g)} & \rightleftharpoons & 2SO_2_{(g)} & + & O_2_{(g)}\\At\ time, t = 0 & 0.5067 &&0&&0 \\ At\ time, t=t_{eq} & -2x &&2x&&x \\----------------&-----&-&-----&-&-----\\Concentration\ at\ equilibrium:- &0.5067-2x&&2x&&x\end{matrix}$

Given:

Equilibrium concentration of O₂ = 0.130 mol

Volume = 1.50 L

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$Molarity=\frac{0.130}{1.50\ L}$

[O₂] = x = 0.0867 M

[SO₂] = 2x = 0.1733 M

[SO₃] = 0.5067-2x = 0.3334 M

The expression for the equilibrium constant is:

$K_c=\frac {[SO_2]^2[O_2]}{[SO_3]^2}$

$K_c=\frac{(0.3334)^2\times 0.0867}{(0.3334)^2}$

$K_c=0.0867$

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4 years ago