## Chemistry (4th Edition)

Published by McGraw-Hill Publishing Company

# Chapter 15 - Questions and Problems - Page 712: 15.70

#### 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.

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