To determine the new initial rate of the reaction when the concentration of H2 is halved, we need to understand the rate law for the reaction. The rate law is generally given by:

$$ \text{Rate} = k [\text{H}_2]^m [\text{NH}_3]^n $$

where $ k $ is the rate constant, and $ m $ and $ n $ are the reaction orders with respect to H2 and NH3, respectively.

Given that the initial rate is 0.830 M/s, we can write:

$$ 0.830 = k [\text{H}_2]^m [\text{NH}_3]^n $$

When the concentration of H2 is halved, the new concentration of H2 is $ \frac{[\text{H}_2]}{2} $. The new rate can be expressed as:

$$ \text{New Rate} = k \left( \frac{[\text{H}_2]}{2} \right)^m [\text{NH}_3]^n $$

This simplifies to:

$$ \text{New Rate} = k \frac{[\text{H}_2]^m}{2^m} [\text{NH}_3]^n $$

$$ \text{New Rate} = \frac{1}{2^m} \times k [\text{H}_2]^m [\text{NH}_3]^n $$

Since $ k [\text{H}_2]^m [\text{NH}_3]^n = 0.830 $:

$$ \text{New Rate} = \frac{1}{2^m} \times 0.830 $$

Without the specific value of $ m $, we cannot determine the exact new rate. However, if we assume the reaction is first-order with respect to H2 (i.e., $ m = 1 $), then:

$$ \text{New Rate} = \frac{1}{2^1} \times 0.830 $$

$$ \text{New Rate} = \frac{1}{2} \times 0.830 $$

$$ \text{New Rate} = 0.415 $$

Therefore, if the reaction is first-order with respect to H2, the new initial rate of the reaction when the concentration of H2 is halved would be 0.415 M/s.

Question

To determine the new initial rate of the reaction when the concentration of H2 is halved, we need to understand the rate law for the reaction. The rate law is generally given by:

$$ \text{Rate} = k [\text{H}_2]^m [\text{NH}_3]^n $$

where $ k $ is the rate constant, and $ m $ and $ n $ are the reaction orders with respect to H2 and NH3, respectively.

Given that the initial rate is 0.830 M/s, we can write:

$$ 0.830 = k [\text{H}_2]^m [\text{NH}_3]^n $$

When the concentration of H2 is halved, the new concentration of H2 is $ \frac{[\text{H}_2]}{2} $. The new rate can be expressed as:

$$ \text{New Rate} = k \left( \frac{[\text{H}_2]}{2} \right)^m [\text{NH}_3]^n $$

This simplifies to:

$$ \text{New Rate} = k \frac{[\text{H}_2]^m}{2^m} [\text{NH}_3]^n $$

$$ \text{New Rate} = \frac{1}{2^m} \times k [\text{H}_2]^m [\text{NH}_3]^n $$

Since $ k [\text{H}_2]^m [\text{NH}_3]^n = 0.830 $:

$$ \text{New Rate} = \frac{1}{2^m} \times 0.830 $$

Without the specific value of $ m $, we cannot determine the exact new rate. However, if we assume the reaction is first-order with respect to H2 (i.e., $ m = 1 $), then:

$$ \text{New Rate} = \frac{1}{2^1} \times 0.830 $$

$$ \text{New Rate} = \frac{1}{2} \times 0.830 $$

$$ \text{New Rate} = 0.415 $$

Therefore, if the reaction is first-order with respect to H2, the new initial rate of the reaction when the concentration of H2 is halved would be 0.415 M/s.

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