Q. 7.76 - An Introduction to General, Organic, and Biological Chemistry

Question

How would each of the following change the rate of the reaction shown here:

$2NO\left(g\right) + 2H_2\left(g\right) \to N_2\left(g\right) + 2H_{2}0\left(g\right)$

a. adding some $NO\left(g\right)$

b. decreasing the temperature

c. removing some $H_2\left(g\right)$

d. adding a catalyst

Step-by-Step Solution

Answer

(a) Rate of reaction moves toward forward .

(b) Rate of reaction moves toward  backward.

(c) Rate of reaction moves toward backward.

(d) Rate of reaction moves toward forward.

Part (a) Step1 : Given Information

We need to find the rate if we adding some NO(g) .

$2NO\left(g\right) + 2H_2\left(g\right) \to N_2\left(g\right) + 2H_{2}0\left(g\right)$

Part (a) Step 2: Explanation

If we add some $NO$, the concentration of the reactant will rise, and the equilibrium reaction will move forward to balance.

Part (b) Step 1: Given Information

We need to find the rate when decreasing the temperature.

Part (b) Step 2: Explanation

If we decrease the temp then by the equilibrium equation:

 $$\begin{gathered} K_p = K_c{\left(RT\right)}^{∆ng} K_c = \left(RT\right)K_p \\ {K}_c \alpha T \end{gathered}$$

Hence if T is decreasing then rate is also decreasing so reaction will move backward.

Part (c) Step 1: Given Information

We need to find the rate when removing some amount of $H_2\left(g\right)$.

Part (c) Step 2: Explanation

If we removing some $H_2\left(g\right)$ then concentration of reactant will be decreasing and then to balanced the equilibrium reaction move backward. 

Part (d) Step 1: Given Information

We need to find the rate when we add a catalyst.

Part (d) Step 2: Explanation

If we add a catalyst then rate of reaction will be increase then reaction rate will be forward.