For the reaction $$\frac{1}{3} N_2(g) + H_2(g) \rightleftharpoons \frac{2}{3} NH_3(g)$$, the magnitude of $K_c$ at 25 °C is: A. 9 and $\Delta H = -30.8$ kJ mol$^{-1}$ B. 213 and $\Delta H = -30.8$ kJ mol$^{-1}$ C. 640 and $\Delta H = -30.8$ kJ mol$^{-1}$ D. 640 and $\Delta H = -92.3$ kJ mol$^{-1}$

SSCE VCE Chemistry Extent of Chemical Reactions: For the reaction $$\frac{1}{3}

For the reaction $$\frac{1}{3} N_2(g) + H_2(g) \rightleftharpoons \frac{2}{3} NH_3(g)$$, the magnitude of $K_c$ at 25 °C is: A - VCE - SSCE Chemistry - Question 14 - 2020 - Paper 1

Question 14

$For-the-reaction--$$\frac{1}{3}-N_2(g)-+-H_2(g)-\rightleftharpoons-\frac{2}{3}-NH_3(g)$$,-the-magnitude-of-$K_c$-at-25-°C-is:--A-VCE-SSCE Chemistry-Question 14-2020-Paper 1.png$

For the reaction $$\frac{1}{3} N_2(g) + H_2(g) \rightleftharpoons \frac{2}{3} NH_3(g)$$, the magnitude of $K_c$ at 25 °C is: A. 9 and $\Delta H = -30.8$ kJ mol$^{-... show full transcript

Worked Solution & Example Answer:For the reaction $$\frac{1}{3} N_2(g) + H_2(g) \rightleftharpoons \frac{2}{3} NH_3(g)$$, the magnitude of $K_c$ at 25 °C is: A - VCE - SSCE Chemistry - Question 14 - 2020 - Paper 1

Step 1

Determine the equilibrium constant and its relation to ΔH

96%

114 rated

Answer

To find the magnitude of the equilibrium constant ( $K_c$ ) at 25 °C, we can refer to the equation that relates $K_c$ and the enthalpy change ( $\Delta H$ ). A general relationship is given by van 't Hoff's equation:

$\ln\left(\frac{K_{c2}}{K_{c1}}\right) = -\frac{\Delta H}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)$

Where:

$R$ is the gas constant, approximately $8.31$ J mol $^{-1}$ K $^{-1}$ .
T is the temperature in Kelvin.

For this problem, we analyze the possible options and their corresponding ΔH values, recognizing that $K_c$ should increase with a decrease in $\Delta H$ since the reaction is exothermic.

Step 2

Evaluate each option

99%

104 rated

Answer

Based on the given options:

Option A: $K_c = 9$ , $\Delta H = -30.8$ kJ mol $^{-1}$
Option B: $K_c = 213$ , $\Delta H = -30.8$ kJ mol $^{-1}$
Option C: $K_c = 640$ , $\Delta H = -30.8$ kJ mol $^{-1}$
Option D: $K_c = 640$ , $\Delta H = -92.3$ kJ mol $^{-1}$

We can rule out options that contradict the principles of thermodynamics. The logical option would be where $K_c$ has a notable increase while remaining consistent with the ΔH values.

Step 3

Select the correct answer

96%

101 rated

Answer

From our evaluation, the most appropriate choice that reflects a sound relationship between the equilibrium constant and enthalpy change is B. 213 and $\Delta H = -30.8$ kJ mol $^{-1}$ .

For the reaction $$\frac{1}{3} N_2(g) + H_2(g) \rightleftharpoons \frac{2}{3} NH_3(g)$$, the magnitude of $K_c$ at 25 °C is: A - VCE - SSCE Chemistry - Question 14 - 2020 - Paper 1

Worked Solution & Example Answer:For the reaction $$\frac{1}{3} N_2(g) + H_2(g) \rightleftharpoons \frac{2}{3} NH_3(g)$$, the magnitude of $K_c$ at 25 °C is: A - VCE - SSCE Chemistry - Question 14 - 2020 - Paper 1

Determine the equilibrium constant and its relation to ΔH

Evaluate each option

Select the correct answer

Join the SSCE students using SimpleStudy...