In the reaction below, ΔH°f is zero for ___________.
Ni(s) + 2CO(g) + 2PF3(g) = Ni(CO)2(PF3)2(l)
Ni(s)
CO(g)
Ni(CO)2(PF3)2(l)
both CO(g) and PF3(g)
PF3(g)
The standard heat of formation for any element in its natural state is zero.
Therefore, ΔH°f for Ni(s) is zero.
Well, let me tell you a chemistry joke. Why was the math book sad? Because it had too many problems! Now, let's talk about the reaction. In this case, ΔH°f is zero for the element in its standard state. Can you guess which one it is?
In the given reaction, ΔH°f is zero for "Ni(s)" and "PF3(g)."
To determine in which component ΔH°f is zero in the given reaction, we need to understand what ΔH°f represents.
ΔH°f (standard enthalpy of formation) is the change in enthalpy that occurs when one mole of a compound is formed from its elements in their standard states. The standard state is the most stable form of an element at a given temperature and pressure.
In the given reaction, we have Ni(s) (solid nickel), CO(g) (gaseous carbon monoxide), PF3(g) (gaseous phosphorus trifluoride), and Ni(CO)2(PF3)2(l) (liquid nickel bis-carbonyl bis-phosphine).
To find out where ΔH°f is zero, we need to determine the standard enthalpy of formation values for these components. ΔH°f is zero for elements in their standard state because they are formed from themselves. However, compounds have non-zero ΔH°f values.
To find ΔH°f values, we can refer to standard enthalpy of formation tables or use thermodynamic data. By consulting these sources, we can determine that ΔH°f is zero for elements in their standard states, such as Ni(s).
Hence, in the given reaction, ΔH°f is zero for Ni(s) (solid nickel).