Houston Texans QB C.J. Stroud was named AFC Offensive Player of the Week, the NFL announced. The honor comes after Stroud’s record-setting performance in the team’s 39-37 last-second victory over the Tampa Bay Buccaneers. This marks Stroud’s first such award in his career after garnering NFL Offensive Rookie of the Month for September.
In Week 9, Stroud completed 30-of-42 pass attempts (71.4 percent) for 470 yards, five touchdowns and zero interceptions for a passer rating of 147.8. In the second half, the rookie signal caller completed 16-of-20 pass attempts for 325 yards, four touchdowns and a perfect passer rating of 158.3. With :46 seconds left and two timeouts, Stroud drove the team down the field, completing all five passes for 75 yards, including a 26-yard dart to WR Tank Dell that setup the ensuing 15-yard game-winning touchdown to Dell with just :06 seconds remaining to complete his first career fourth-quarter comeback and a Texans victory.
Stroud’s memorable day broke numerous records, including setting a new single-game passing mark for most passing yards by a rookie in NFL history. Additionally, his 147.8 passer rating ranks as the highest rating in NFL annals by a rookie quarterback with at least 30 pass attempts. He became just the sixth player in NFL history to record at least 450 passing yards and five touchdown passes with no interceptions in a game, while standing as the youngest player ever to surpass the 450-yard barrier.
On the season, Stroud has started in all nine contests for the Texans, completing 173-of-279 pass attempts for 2,270 yards, 14 touchdowns to just one interception for a 102.9 passer rating. His 14-1 touchdown to interception ratio ranks as the sixth-best mark in league history through the first nine weeks of the season.
The accolade marks the 53rd time a Texan has been named Player of the Week and the 18th time an offensive player has won the award in franchise history.