Question

For the reaction
$H_2PO_2^- (aq) + OH^- (aq) → HPO_3^{2-} (aq) + H_2(g)$
the rate expression is $k[H_2PO_2 ][OH ]^2$. If the concentration of $H_2PO_2^-$ is

• A. tripled
• B. halved
• C. doubled
• D. unchanged

Answer 1

The Correct Option is C

$\text{1. Rate Expression}$

The general rate expression for the reaction is:

$$\text{Rate}_1 = k[\text{H}_2\text{PO}_2^-][\text{OH}^-]^2$$

$\text{2. Effect of Doubling } [\text{H}_2\text{PO}_2^-]$

We introduce the change:

The concentration of $\text{H}_2\text{PO}_2^-$ is doubled: $[\text{H}_2\text{PO}_2^-]_{\text{new}} = 2 \times [\text{H}_2\text{PO}_2^-]_{\text{original}}$

The concentration of $\text{OH}^-$ is kept constant: $[\text{OH}^-]_{\text{new}} = [\text{OH}^-]_{\text{original}}$

The new reaction rate ($\text{Rate}_2$) is calculated by substituting these new concentrations into the rate expression:

$$\text{Rate}_2 = k (2[\text{H}_2\text{PO}_2^-])([\text{OH}^-])^2$$

$\text{3. Comparison of Rates}$

To find how the rate is affected, we compare the new rate ($\text{Rate}_2$) to the original rate ($\text{Rate}_1$):

$$\frac{\text{Rate}_2}{\text{Rate}_1} = \frac{k (2[\text{H}_2\text{PO}_2^-])([\text{OH}^-])^2}{k[\text{H}_2\text{PO}_2^-][\text{OH}^-]^2}$$

The terms $k$, $[\text{H}_2\text{PO}_2^-]$, and $[\text{OH}^-]^2$ cancel out:

$$\frac{\text{Rate}_2}{\text{Rate}_1} = 2$$

$$\text{Rate}_2 = 2 \times \text{Rate}_1$$

$\text{Conclusion}$

The reaction is first-order with respect to $\text{H}_2\text{PO}_2^-$ (because its exponent in the rate law is $1$). Therefore, if the concentration of $\text{H}_2\text{PO}_2^-$ is doubled while keeping the $\text{OH}^-$ concentration constant, the reaction rate will be doubled.

$$\text{The reaction rate is } \mathbf{\text{doubled}}$$