Which of the following reactions will result in a reduced total pressure?
a.
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
c.
2HI(g) → H2(g) + I2(g)
b.
2N2O(g) → 2N2(g) + O2(g)
d.
2H2(g) + O2(g) → 2H2O()
To determine which of the reactions will result in a reduced total pressure, we need to consider the stoichiometry of the reactants and products. The key concept to keep in mind is the ideal gas law, which states that pressure is directly proportional to the number of moles of gas present.
In reaction (a):
CH4(g) + 2O2(g) -> CO2(g) + 2H2O(g)
The total number of moles of gas on the left side is 1 + 2 = 3 moles, and the total number of moles on the right side is 1 + 2 = 3 moles. So, the total pressure does not change in this reaction.
In reaction (b):
2N2O(g) -> 2N2(g) + O2(g)
The total number of moles of gas on the left side is 2, and the total number of moles on the right side is 2 + 1 = 3. Therefore, the total pressure increases in this reaction.
In reaction (c):
2HI(g) -> H2(g) + I2(g)
The total number of moles of gas on the left side is 2, and the total number of moles on the right side is 1 + 1 = 2. Thus, the total pressure decreases in this reaction.
In reaction (d):
2H2(g) + O2(g) -> 2H2O(g)
The total number of moles of gas on the left side is 2 + 1 = 3, and the total number of moles on the right side is 2. Therefore, the total pressure decreases in this reaction.
Based on this analysis, reaction (c) - 2HI(g) -> H2(g) + I2(g) - will result in a reduced total pressure.
To determine which of the given reactions will result in a reduced total pressure, we need to examine the number of moles of gaseous reactants and products.
a. CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
In this reaction, there are 3 moles of gaseous reactants (1 mole of CH4 and 2 moles of O2) and 3 moles of gaseous products (1 mole of CO2 and 2 moles of H2O). The total number of moles before and after the reaction is the same, so the total pressure will not be reduced by this reaction.
c. 2HI(g) → H2(g) + I2(g)
In this reaction, there are 2 moles of gaseous reactants (2 moles of HI) and 3 moles of gaseous products (1 mole of H2 and 1 mole of I2). The total number of moles before the reaction is greater than the total number of moles after the reaction, so the total pressure will be reduced by this reaction.
b. 2N2O(g) → 2N2(g) + O2(g)
In this reaction, there are 2 moles of gaseous reactants (2 moles of N2O) and 3 moles of gaseous products (2 moles of N2 and 1 mole of O2). The total number of moles before the reaction is greater than the total number of moles after the reaction, so the total pressure will be reduced by this reaction.
d. 2H2(g) + O2(g) → 2H2O(g)
In this reaction, there are 3 moles of gaseous reactants (2 moles of H2 and 1 mole of O2) and 2 moles of gaseous products (2 moles of H2O). The total number of moles before the reaction is greater than the total number of moles after the reaction, so the total pressure will be reduced by this reaction.
Therefore, the reactions that will result in a reduced total pressure are c. 2HI(g) → H2(g) + I2(g) and b. 2N2O(g) → 2N2(g) + O2(g).