Question

deltaf U of formation of CH4 at certain temperature is -393kj mole the value of deltafH is

• A. < Δ U
• B.

  Δ U

• C. = Δ U

Answer 1

Answer: (A)

< Δ U

Answer 2

To determine the relationship between $\Delta_f H$ and $\Delta_f U$ for the formation of CH$_4$, we use the equation:

$\Delta H = \Delta U + \Delta n_g RT$

First, write the balanced chemical equation for the formation of CH$_4$ from its elements in their standard states:

$C(s, graphite) + 2H_2(g) \rightarrow CH_4(g)$

Next, calculate the change in the number of moles of gaseous species, $\Delta n_g$:

$\Delta n_g = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants})$

From the equation:

Moles of gaseous products = 1 (for CH$_4$(g)) Moles of gaseous reactants = 2 (for 2H$_2$(g))

$\Delta n_g = 1 - 2 = -1$

Now, substitute this value into the relationship between $\Delta H$ and $\Delta U$:

$\Delta_f H = \Delta_f U + (-1)RT$ $\Delta_f H = \Delta_f U - RT$

Given that $\Delta_f U = -393 \text{ kJ/mol}$.

So, $\Delta_f H = -393 - RT$.

Since R (the ideal gas constant) and T (absolute temperature in Kelvin) are always positive values, their product $RT$ will also be a positive value.

Subtracting a positive value ($RT$) from $\Delta_f U$ will result in a value that is algebraically smaller than $\Delta_f U$.

Therefore, $\Delta_f H < \Delta_f U$.