What would Qc be if the volume of the container decreased by a factor of 1.774 for the following reaction?
PH3(g) + BCl3(g) → PH3BCl3(s)
KC = 534.8
at 353.0oC
If the volume decreases (by say a factor of 2) that will double the (PH3) and double (BCl3). For a factor of 1.774 concn will be increased for EACH by 1.774. Remember the solid doesn't appear in Kc.
Kc = 1/(PH3)(BCl3)
So Kc > Q = 534.8/(1.774)(1.774)
To determine the change in heat (Qc) when the volume of the container decreases, we need to use the equation:
Qc = Δn * R * T
Where:
- Qc is the change in heat
- Δn is the change in the number of moles of gas between the initial and final states
- R is the ideal gas constant
- T is the temperature in Kelvin
Now, let's calculate the change in moles of gas (Δn). The balanced equation shows that 1 mol of PH3(g) and 1 mol of BCl3(g) react to form 1 mol of PH3BCl3(s). Therefore, the change in moles (Δn) is zero, as the number of moles of gas remains the same.
Next, we'll need to convert the temperature from Celsius to Kelvin by using the formula:
T(K) = T(°C) + 273.15
Given that the temperature is 353.0 °C, we can convert it to Kelvin:
T(K) = 353.0 + 273.15 = 626.15 K
With the values of Δn = 0, R = 8.314 J/(mol·K), and T = 626.15 K, we can calculate Qc using the formula mentioned earlier:
Qc = Δn * R * T
= 0 * 8.314 * 626.15
= 0
Therefore, the change in heat (Qc) when the volume of the container decreases by a factor of 1.774 is zero.