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Fofino [41]
4 years ago
15

The reaction 2a + b --> c + d has an activation energy of 80.0 kj/mol. at 320°c, the rate constant k = 1.80 x 10-2 l mol-1 s-

1. what would be the value of k at 450°c?
Chemistry
1 answer:
Naddik [55]4 years ago
5 0
To determine the k for the second condition, we use the Arrhenius equation which relates the rates of reaction at different temperatures. We do as follows:

ln k1/k2 = E / R (1/T2 - 1/T1) where E is the activation energy and R universal gas constant.

ln 1.80x10^-2 / k2 = 80000 / 8.314 ( 1/723.15 - 1/593.15)

k2 = 0.3325 L / mol-s
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How many atoms do the elements of sodium hypochlorite have?
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The chemical formula is NaClO and consists of one atom of sodium (Na), one atom of chlorine (Cl) and one atom of oxygen (O).

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3 years ago
A mixture of 0.307 M Cl 2 , 0.465 M F 2 , and 0.706 M ClF is enclosed in a vessel and heated to 2500 K . Cl 2 ( g ) + F 2 ( g )
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<u>Answer:</u> The equilibrium concentration of chlorine gas, fluorine gas and ClF gas is 0.159 M, 0.317 M and 1.002 M respectively.

<u>Explanation:</u>

We are given:

Initial concentration of chlorine gas = 0.307 M

Initial concentration of fluorine gas = 0.465 M

Initial concentration of ClF gas = 0.706 M

The given chemical equation follows:

                            Cl_2(g)+F_2(g)\rightleftharpoons 2ClF(g)

<u>Initial:</u>                  0.307       0.465       0.706

<u>At eqllm:</u>           0.307-x    0.465-x       0.706+2x

The expression of K_c for above equation follows:

K_c=\frac{[ClF]^2}{[Cl_2][F_2]}

We are given:

K_c=20.0

Putting values in above equation, we get:

20.0=\frac{(0.706+2x)^2}{(0.307-x)(0.465-x)}\\\\x=0.148,0.993

Neglecting the value of x = 0.993 because the equilibrium concentrations of chlorine and fluorine gases will become negative, which is not possible

So, equilibrium concentration of chlorine gas = (0.307 - x) = [0.307 - 0.148] = 0.159 M

Equilibrium concentration of fluorine gas = (0.465 - x) = [0.465 - 0.148] = 0.317 M

Equilibrium concentration of ClF gas = (0.706 + 2x) = [0.706 + 2(0.148)] = 1.002 M

Hence, the equilibrium concentration of chlorine gas, fluorine gas and ClF gas is 0.159 M, 0.317 M and 1.002 M respectively.

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3 years ago
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