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3 years ago

Hydrogen is manufactured on an industrial scale by this sequence of reactions:

Chemistry

1 answer:

hichkok12 [17]3 years ago

6 0

Answer:

$K=K_1*K_2\\\\K=\frac{[H_2]^3[CO_2][H_2]}{[CH_4][H_2O][H_2O]}$

Explanation:

Hello there!

In this case, for the given chemical reaction, it turns out firstly necessary to write the equilibrium expression for both reactions 1 and 2:

$K_1=\frac{[CO][H_2]^3}{[CH_4][H_2O]} \\\\K_2=\frac{[CO_2][H_2]}{[CO][H_2O]}$

Now, when we combine them to get the overall expression, we infer these two are multiplied to get:

$K=K_1*K_2\\\\K=\frac{[CO][H_2]^3}{[CH_4][H_2O]} *\frac{[CO_2][H_2]}{[CO][H_2O]}\\\\K=\frac{[H_2]^3[CO_2][H_2]}{[CH_4][H_2O][H_2O]}$

Regards!

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3 years ago

The empirical formula of a compound is CH. At 200 degree C, 0.145 g of this compound occupies 97.2 mL at a pressure of 0.74 atm.

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Answer:

The molecular formula = $C_{6}H_{6}$

Explanation:

Given that:

Mass of compound, m = 0.145 g

Temperature = 200 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (200 + 273.15) K = 473.15 K

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Considering,

$n=\frac{m}{M}$

Using ideal gas equation as:

$PV=\frac{m}{M}RT$

where,

P is the pressure

V is the volume

m is the mass of the gas

M is the molar mass of the gas

T is the temperature

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the values in the above equation as:-

$0.74\times 0.0972=\frac{0.145}{M}\times 0.0821\times 473.15$

$M=78.31\ g/mol$

The empirical formula is = $CH$

Molecular formulas is the actual number of atoms of each element in the compound while empirical formulas is the simplest or reduced ratio of the elements in the compound.

Thus,

Molecular mass = n × Empirical mass

Where, n is any positive number from 1, 2, 3...

Mass from the Empirical formula = 12 + 1 = 13 g/mol

Molar mass = 78.31 g/mol

So,

Molecular mass = n × Empirical mass

78.31 = n × 13

⇒ n ≅ 6

The molecular formula = $C_{6}H_{6}$

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