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3 years ago

What is △n for the following equation in relating Kc to Kp?

Chemistry

1 answer:

Nimfa-mama [501]3 years ago

7 0

Answer:

-1

Explanation:

The relation between Kp and Kc is given below:

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

Where,

Kp is the pressure equilibrium constant

Kc is the molar equilibrium constant

R is gas constant , 0.082057 L atm.mol⁻¹K⁻¹

T is the temperature in Kelvins

Δn = (No. of moles of gaseous products)-(No. of moles of gaseous reactants)

For the first equilibrium reaction:

$2Na_{(s)}+2H_2O_{(l)}\rightleftharpoons 2NaOH_{(aq)}+2H_2_{(g)}$

Δn = (No. of moles of gaseous products)-(No. of moles of gaseous reactants) = (2+1)-(2+2) = -1

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At constant pressure, which of these systems do work on the surroundings? A ( s ) + B ( s ) ⟶ C ( g ) A(s)+B(s)⟶C(g) 2 A ( g ) +

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Correct question:

At constant pressure, which of these systems do work on the surroundings?

(a) A ( s ) + B ( s ) ⟶ C ( g )

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(d) 2 A ( g ) + 2 B ( g ) ⟶ 3 C ( g )

Answer:

(a) A ( s ) + B ( s ) ⟶ C ( g )

(b) 2 A ( g ) + 2 B ( g ) ⟶ 5 C ( g )

Explanation:

Work done by a system on the surroundings at a constant pressure is given as;

W = -PΔV

Where;

ΔV is gas expansion, that is final volume of the gas minus initial volume of the gas must be greater than zero.

Part (a)

A ( s ) + B ( s ) ⟶ C ( g )

ΔV = 1 - (0) = 1 (expansion)

Part (b)

2 A ( g ) + 2 B ( g ) ⟶ 5 C ( g )

ΔV = 5 - ( 2+ 2) = 1 (expansion)

Part (c)

A ( g ) + B ( g ) ⟶ C ( g )

ΔV = 1 - ( 1 + 1) = -1 (compression)

Part (d)

2 A ( g ) + 2 B ( g ) ⟶ 3 C ( g )

ΔV = 3 - ( 4) = -1 (compression)

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(a) A ( s ) + B ( s ) ⟶ C ( g )

(b) 2 A ( g ) + 2 B ( g ) ⟶ 5 C ( g )

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3 years ago

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Explanation:

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12345 [234]

1) 385 J

2) 450 J

Explanation:

The amount of energy that must be absorbed by a certain substance in order to increase its temperature by $\Delta T$ is given by the equation:

$Q=mC\Delta T$

where

m is the mass of the substance

C is its specific heat capacity

$\Delta T$ is the increase in temperature of the substance

For the block of copper in this problem, we have:

m = 10 g is the mass

$C=0.385 J/gK$ is the specific heat capacity of copper

$\Delta T=400-300 = 100 K$ is the change in temperature

So, the energy absorbed by the block of copper is

$Q=(10)(0.385)(100)=385 J$

Similarly for the block of iron, the energy absorbed by the iron is given by

$Q=mC\Delta T$

where

m is the mass of the block of iron

C is its specific heat capacity of iron

$\Delta T$ is the increase in temperature of the block

Here we have:

m = 10 g is the mass of the block

$C=0.450 J/gK$ is the specific heat capacity of iron

$\Delta T=400-300 = 100 K$ is the change in temperature

So, the energy absorbed by the block of iron is

$Q=(10)(0.450)(100)=450 J$

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