Answer
T = 200 $^oC$:
$K_C =59.0$
$K_p = 2.29 \times 10^3$
T = 300 $^o C$
$K_C = 3.40$
$K_p = 1.60 \times 10^2$
T = 400 $^o C$
$K_C = 2.10$
$K_p = 1.16 \times 10^2$
The reaction is exothermic.
Work Step by Step
T = 200:
1. Write the equilibrium constant expression:
- The exponent of each concentration is equal to its balance coefficient.
$$K_C = \frac{[Products]}{[Reactants]} = \frac{[ B ] ^{ 2 }}{[ A ]}$$
$K_p = K_c(RT)^{\Delta n} = K_c(0.08206*T)^{2 -1 } = K_c(0.08206T)$
2. Substitute the values and calculate the constant value:
$$K_C = \frac{( 0.843 )^{ 2 }}{( 0.0125 )} = 59.0$$
$K_p = (59.0)(0.08206)(200 + 273.15) = 2.29 \times 10^3$
T = 300:
2. Substitute the values and calculate the constant value:
$$K_C = \frac{( 0.764 )^{ 2 }}{( 0.171 )} = 3.40$$
$K_p = (3.40)(0.08206)(300 + 273.15) = 1.60 \times 10^2$
T = 400:
2. Substitute the values and calculate the constant value:
$$K_C = \frac{( 0.724 )^{ 2 }}{( 0.250 )} = 2.10$$
$K_p = (2.10)(0.08206)(400 + 273.15) = 1.16 \times 10^2$
$K_c$ decreases as the temperature increases; therefore, the reaction is exothermic.