A mixture can be classified as a solution suspension or colloid based on the?
1 answer:
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Answer : The value of equilibrium constant for this reaction at 262.0 K is
Explanation :
As we know that,
where,
= standard Gibbs free energy = ?
= standard enthalpy = -45.6 kJ = -45600 J
= standard entropy = -125.7 J/K
T = temperature of reaction = 262.0 K
Now put all the given values in the above formula, we get:
The relation between the equilibrium constant and standard Gibbs free energy is:
where,
= standard Gibbs free energy = -12666.6 J
R = gas constant = 8.314 J/K.mol
T = temperature = 262.0 K
K = equilibrium constant = ?
Now put all the given values in the above formula, we get:
Therefore, the value of equilibrium constant for this reaction at 262.0 K is
n = 0.100 moles
At STP (standard temperature pressure), where gas is collected, the temperature is 0°C and the pressure is 1 atm.
The ideal gas equation indicates that
Group of answer choices .The volume occupied by mole of a given gas at a given temperature and pressure is expressed as the gas's molar volume.
The most typical illustration is the molar volume of a gas at STP, which is equal to for
mole of any ideal gas at a temperature of
Kand a pressure of
atm.
Hence, the volume of the gas is .
A unit is stated to have a temperature of absolute zero and an atmospheric pressure of one atmosphere, or
atm, at standard temperature and pressure. Additionally, at STP, a mole of any gas takes up
of space. Keep in mind that this idea only applies to gases.
For experimental measurements to be established under standard conditions that allow for comparisons between various sets of data, standard temperature and pressure must be met.
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